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Let $a, b, c > 0$ with $a + b + c = 3$. Prove that $$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$

This question was posted recently, closed and then deleted, due to missing of contexts etc.

By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.

My strategy is to split into many cases.

WLOG, assume that $a \ge b \ge c$.

If $a \ge 8/5$, true.

If $a \le 10/7$, let $f(x) = x^{4/x} - 1 - 4(x - 1)$. We have $f(x) \ge 0$ for all $x \in (0, 10/7)$.

If $10/7 < a < 8/5$ and $b \ge 4/5$, true.

(I stopped here since this approach is ugly. Actually, the proof of $x^{4/x} - 1 - 4(x - 1) \ge 0$ for all $x\in (0, 10/7)$ is complicated.)

I hope to see nice proofs.

River Li
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    I've found that $f(x) = x^{1/x}$ has its maximum value at $x = e$, but this doesn't help as much for a lower bound. – Dan Jul 29 '22 at 00:24
  • Maybe Lagrange multiplier can find the minimum of that smooth function. – Mason Jul 29 '22 at 01:22
  • @Mason I think so. Also, I prefer non-LM approaches. LM is something like "Break a butterfly on the wheel". – River Li Jul 29 '22 at 02:56
  • Can you see the deleted post? It seems that a user wrote an answer using the Bernoulli inequality. While I haven't verified it, it goes as follows : write $a=x+1,b=y+1,c=z+1$, and use $(x+1)^{\frac 4{x+1}} \geq 1+\frac{4x}{x+1}$ (and likewise for the others) using Bernoulli's inequality, and sum up to finish. I think there is something wrong after the summing up, personally, but I'll leave this up to indicate that Bernoulli might be a way out. – Sarvesh Ravichandran Iyer Jul 29 '22 at 06:03
  • @SarveshRavichandranIyer Yes, I can see that answer. The author deleted it (I think there is something wrong). – River Li Jul 29 '22 at 06:06
  • @RiverLi Looking back, there probably is. There is an assertion that $\frac{4x}{x+1} + \frac{4y}{y+1} + \frac{4z}{z+1} \geq 0$, which reduces to $\frac{x}{x+1}+\frac{y}{y+1} + \frac{z}{z+1} \geq 0$ whenever $x+y+z=0$. That, I'm not sure it's true and would cause the proof to collapse. Perhaps one can go one level "deeper" in Bernoulli's inequality, maybe. – Sarvesh Ravichandran Iyer Jul 29 '22 at 06:09
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    @SarveshRavichandranIyer I believe Bernoulli's inequality plays some role here. – River Li Jul 29 '22 at 06:13
  • grotex, Crazy_littleboy, fjw.., @HN_NH proposed so many hard problems. –  Jul 29 '22 at 10:54
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    @ĐặngHảiĐăng Yes, some of those problems are still open in some sense. – River Li Jul 29 '22 at 12:00
  • Trivially, if $a = b = c = 1$, then $f(a) + f(b) + f(c) = 3$. So the challenge is to prove that that is the minimum sum on the given domain plane. – Dan Jul 29 '22 at 16:52
  • You write: "Actually, the proof of $x^{4/x} - 1 - 4(x - 1) \ge 0$ for all $x\in (0, 10/7)$ is complicated." It can be done with calculus, see my answer below for details. – Andreas Aug 04 '22 at 08:13
  • @Andreas I think that my proof and your proof for $x^{4/x} \ge 4x - 3$ on some interval $(0, c)$ using calculus are both complicated. In contrast, using Bernoulli, it is one line: $$x^{4/x} = (x^{\frac{4}{3x}})^3 \ge \Big(1 + (x - 1)\cdot \frac{4}{3x}\Big)^3 = 4x - 3 + \frac{4(52x - 27x^2 - 16)(x - 1)^2}{27x^3} \ge 4x - 3.$$ – River Li Aug 04 '22 at 09:33
  • Bernoulli, in the form which you use, only holds for $\frac{4}{3x} \ge 1$, i.e. $x \le \frac{4}{3}$, hence it does only prove the desired result for $1 \le x \le \frac{4}{3} \sim 1.333$. But we want it for $1 \le x \le \frac{10}{7} \sim 1.429$, so this is not a one-liner. – Andreas Aug 04 '22 at 11:40
  • @Andreas I split into $(0, 4/3)$ and $(4/3, c)$ which are both proved by Bernoulli. This is much simpler and elementary than taking derivative and second derivatives. – River Li Aug 04 '22 at 12:14

5 Answers5

4

Before the actual proof, let's start with a discussion of $f(x) = x^{4/x}$ which is needed later.

We have $ df(x) /dx = 4 x^{4/x - 2} (1 - \log(x))$ so $f(x)$ is rising for all $0 < x < e$ which is all we need to consider below. Further, $$ d^2f(x) /dx^2 = 4 x^{4/x - 4} (-3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4)$$ To interpret this result for positive and negative ranges, consider $ h(x) = -3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4$. Clearly, $h(x) {{(x\to 0) } \atop{\longrightarrow}} \infty $. Further, $$ h'(x) = (-x + 2 (x + 4) \log(x) - 8)/x $$ which remains negative for at least $0 < x < 2$. So if we have a value where $d^2f(x) /dx^2$ changes sign in the range $0 < x < 2$, this will be the only sign change there. The sign change indeed occurs near $x= 1.125$.

We can now describe the behavior of $f(x) = x^{4/x}$ for $0 < x < 2$. It is rising, it is convex for $0 < x < x_0 $ and concave for $ x_0 < x < 2$ where $x_0≈1.125$ is the unique real solution of $h(x)=0$ on $(0,2)$.

Now consider $g(x) = -3 + 4x$. Note $g(x)$ touches $f(x) = x^{4/x}$ at $x=1$, since $g(1) =f(1) = 1$ and $g'(1) = f'(1) = 4$. Since $x^{4/x}$ is convex at $x=1$ and turns concave at $x \simeq 1.125$, we have an "S-shape" and hence, there is at most one intersection $g(x) = f(x)$ in the range $1.125 < x < 2$. By inspection we have that this occurs near $\bar{x} = 1.429$ . So $x^{4/x} \ge g(x)$ for $x < \bar{x} = 1.429$ and hence, for $a,b,c < \bar{x}$, $a^{4/a} + b^{4/b} + c^{4/c} \ge -9 + 4(a+b+c) =3$.

As OP already noted, only $1.52 > a \ge b \ge c$ must then be considered (since $1.52^{4/1.52} > 3$) . Now if $1.52 > a > \bar{x} = 1.429$, then $a^{4/a} \ge \bar{x}^{4/\bar{x}} \simeq 2.71$ and $b +c \ge 3 - 1.52 = 1.48$. Hence it remains to be shown that for $b +c = 1.48$, that $b^{4/b} + c^{4/c} \ge 3 - 2.71 = 0.29$. This can be established by direct calculation, since indeed $b^{4/b} + (1.48-b)^{4/(1.48-b)} > 0.29$ for all $0 \le b \le 1.48$. $\qquad \Box$

Andreas
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  • Thanks for your answer. For a complete proof, I think the details should be given. For example, it might be difficult (or complicated) to prove that $x^{4/x} \ge 1 + 4(x-1)$ on $(0, 1.429)$. Also, even the convexity of $x^{4/x}$ is not easy. – River Li Aug 02 '22 at 00:14
  • Dear River Li, I augmented by starting with a discussion of $f(x) = x^{4/x}$. – Andreas Aug 03 '22 at 16:38
  • Thanks. The following statement is not accurate: "it is convex for $0 < x < 1.125$ and concave for $1.125 < x < 2$." $1.125$ is not an exact value. It should be $0 < x < x_0$ and $x_0 < x < 2$ where $x_0 \approx 1.125$ is the unique real solution of $h(x) = 0$ on $(0, 2)$. Also, rigorously, your statement "S-shape and hence there is at most one intersection" is the property of concave function, say $f(x) \ge \min(f(a), f(b))$ for a concave function $f(x)$ on $[a, b]$. – River Li Aug 03 '22 at 23:29
  • Although your whole proof with all details will be complicated, it is a proof. (+1). – River Li Aug 03 '22 at 23:31
  • I used your rigorous formulation "0<x<x0 and x0<x<2 where x0≈1.125 is the unique real solution of h(x)=0 on (0,2)" in the main text, thanks. W.r.t. the S-shape, there is s misconception on your side. The argument is as follows: as g(x) is a tangent line to a rising function f(x) which is convex at the touching point x=1, we have f(x) > g(x) initially for x>1 as f(x) moves further up from g(x). Now, as we know that f(x) turns concave at x0≈1.125 > 1, f(x) will again turn towards g(x) for x > x0, which gives the S-shape. (continued) – Andreas Aug 04 '22 at 08:01
  • (countinued) Generally, this concavity for x> x0 can result in 0 or 1 intersections of f(x) and g(x) at some x1> x0. Hence, if we find such an intersection x1 in the interval (x0,2), then f(x) > g(x) for (1,x1). This was used here, with $x1> \bar{x} = 1.429$. Since by calculation $f(\bar{x}) > g(\bar{x})$ , f(x) > g(x) definitely holds for $x \in (1,\bar{x})$. – Andreas Aug 04 '22 at 08:07
  • Your explanation is OK. Alternatively, we can say: Since $h(x) = f(x) - g(x)$ is concave on $[x_0, 2]$, and $f(x_0) \ge f(1) = 0$ and $f(1.429) \ge 0$, using the fact that $h(x) \ge \min(h(x_0), h(1.429)) $, we obtain $h(x) \ge 0$ on $[x_0, 1.429]$. – River Li Aug 04 '22 at 09:16
  • It seems someone downvoted you. It is not me. – River Li Aug 05 '22 at 01:31
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Remark: Here is a proof without calculus e.g. derivative and convexity (except for Bernoulli inequality).

Proof.

WLOG, assume that $a \ge b \ge c$. Clearly, $a \ge 1$ and $c \le 1$.

We split into five cases.

Case 1: $a \ge 2$

Using Bernoulli, we have $$a^{4/a} \ge 1 + (a - 1)\cdot \frac4a = 3 + \frac{2(a - 2)}{a} \ge 3.$$

Case 2: $8/5 \le a < 2$

Using Bernoulli, we have $a^{2/a} \ge 1 + (a - 1)\cdot \frac{2}{a} > 0$. Thus, we have $$a^{4/a} = (a^{2/a})^2 \ge \left(1 + (a - 1)\cdot \frac2a\right)^2 = 3 + \frac{2(3a^2 - 6a + 2)}{a^2} \ge 3.$$

Case 3: $4/3 < a < 8/5$ and $c \le 1/2$

Using Bernoulli inequality, we have \begin{align*} a^{4/a} + b^{4/b} &\ge 1 + (a - 1)\cdot \frac4a + 1 + (b - 1)\cdot \frac4b\\[5pt] &= 10 - \frac4a - \frac4b\\[5pt] &= 10 - \frac4a - \frac{4}{3 - a - c}\\[5pt] &\ge 10 - \frac4a - \frac{4}{3 - a - 1/2}\\[5pt] &= 3 + \frac{35a - 14a^2 - 20}{a(5 - 2a)}\\ &> 3. \end{align*}

Case 4: $4/3 < a < 8/5$ and $c > 1/2$

Using Bernoulli, we have $$a^{2/a} \ge 1 + (a - 1)\cdot \frac2a = 3 - \frac2a > 0, $$ and $$b^{2/b} \ge 1 + (b - 1)\cdot \frac2b = 3 - \frac2b > 0$$ and $$c^{1/c} \ge 1 + (c - 1)\cdot \frac1c = 2 - \frac1c > 0.$$

It suffices to prove that $$\left(3 - \frac2a\right)^2 + \left(3 - \frac2b\right)^2 + \left(2 - \frac1c\right)^4 \ge 3.$$

Let $p = a + b$ and $q = ab$. We have \begin{align*} \left(3 - \frac2a\right)^2 + \left(3 - \frac2b\right)^2 &= 18 - \frac{12p - 8}{q} + \frac{4(p^2 - 4q)}{q^2}\\ &\ge 18 - \frac{12p - 8}{q} + 4(p^2 - 4q)\left(\frac{2}{q} - 1\right)\\ &= 16q + \frac{8p^2 - 12p + 8}{q} - 4p^2 - 14\\ &\ge 2\sqrt{16q \cdot \frac{8p^2 - 12p + 8}{q}} - 4p^2 - 14 \\ &= 16\sqrt{2p^2 - 3p + 2} - 4p^2 - 14 \end{align*} where we have used $\frac{1}{q^2} \ge \frac{2}{q} - 1$ and AM-GM.

From $a > 4/3$, we have $c \le \frac{3 - 4/3}{2} = \frac56$.

It suffices to prove that, for all $c\in (1/2, 5/6]$, $$16\sqrt{2(3-c)^2 - 3(3-c) + 2} - 4(3-c)^2 - 14 + \left(2 - \frac1c\right)^4 \ge 3$$ which is true.

Case 5: $a \le 4/3$

Fact 1: $x^{4/x} \ge 4x - 3$ on $(0, 4/3]$.
(The proof is given at the end.)

By Fact 1, we have $$a^{4/a} + b^{4/b} + c^{4/c} \ge 4(a + b + c) - 9 = 3.$$

We are done.

Proof of Fact 1:

We only need to prove the case $x \in (3/4, 4/3]$.

Using Bernoulli, we have $$x^{4/x} = (x^{\frac{4}{3x}})^3 \ge \Big(1 + (x - 1)\cdot \frac{4}{3x}\Big)^3 = 4x - 3 + \frac{4(52x - 27x^2 - 16)(x - 1)^2}{27x^3} \ge 4x - 3.$$

We are done.

River Li
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    Nice. But your original question (earlier text version) asked for sulutions with as few cases as possible. Now this is just what happened.... – Andreas Aug 04 '22 at 08:11
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    @ĐặngHảiĐăng About 3.3 due to arqady@AoPS in the link I gave in OP. – River Li Aug 04 '22 at 09:19
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    @Andreas However, now, the proof is more elementary, without using calculus (derivative, convexity, Lagrange Multiplier etc.). Actually, I think simple and elementary proofs are more important. For example, I hardly use Lagrange Multiplier for inequalities (particularly the contest inequalities). LM is something like "Break a butterfly on the wheel". – River Li Aug 04 '22 at 09:31
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    @Andreas In other words, if the proof is more elementary, even with more cases, it is fine. My original proof is complicated with more cases, it is not good. – River Li Aug 04 '22 at 09:38
  • @Andreas We split into several cases which are all proved using Bernoulli. Actually, Case 1 and 2 can be combined if calculus is allowed i.e., the case $a \ge 8/5$. I pursue more elementary proofs which are kind to high school students, although my current proof is not nice. – River Li Aug 04 '22 at 12:39
  • @Andreas It is a pity there are always strange downvotes. – River Li Aug 05 '22 at 08:36
  • Although Andreas has a pretty nice solution, I still prefer this one, +1 – V.S.e.H. Aug 05 '22 at 22:53
  • @V.S.e.H. Thanks. – River Li Aug 05 '22 at 23:47
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Not an answer just a remark :

Playing with Vasc's lemma 7.1 see (https://www.isr-publications.com/jnsa/articles-1563-proofs-of-three-open-inequalities-with-power-exponential-functions) we have the inequality for $1\leq x \leq 2$ and $a=x,c=\frac{2}{x}$ then :

$$\left((1-c)^{2}+ac(2-c)-ac(1-c)\ln a\right)^{2}\leq f(x)=x^{\frac{4}{x}}$$

I think it's pretty sharp to show the inequality even if now it seems a bit harder for the eyes .

On the other hand it seems that $f(x)$ is convex on $(0,1]$ so we can use the tangent line method .

I haven't yet an idea to find a lower bound on $(2,3)$.(No need in fact).

Edit we have using a simple bound on logarithm ($1\leq x \leq 2$,$a=x,c=\frac{2}{x}$):

$$\left((1-c)^{2}+ac(2-c)-ac(1-c)\cdot\frac{2\left(a-1\right)}{a+1}\right)^{2}\leq x^{\frac{4}{x}}$$

We can also use the RiverLi's bound or something else like for $x\in (0,1]$ :

$$x^{\frac{4}{x}}\geq g(x)=x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)$$

And obviously $0<x^{\frac{4}{x}}$.

It works as well .

We can also use Jensen's inequality because it's not hard to show that $g(x)$ is convex on $[1/3,1]$ and we have a polynomial inequality for $1\leq x \leq 2$:

$$2g\left(\frac{3-x}{2}\right)+f\left(x\right)\geq 3$$

We can also show that $f(x)$ is convex on $(0,9/8]$ remarking that :

$$2\cdot1.125^{\frac{4}{1.125}}>3$$

And use Jensen's inequality .

Remains to show the case $a\in(1,1.125),b\in(0,0.875),c\in(1.125,2)$ and for that use Jensen's inequality on the variable $a,b$ and we have with a such constraint :

$$2f\left(\frac{a+b}{2}\right)+f\left(c\right)\geq 3$$

And use the bounds above .

We are done for this sketch of proof .

Miss and Mister cassoulet char
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Using Vasc's paper and @Andreas have already shown that $f\left(x\right)=x^{\frac{4}{x}}$ is convex for $x\in(0,1]$ .So using WLCF-Theorem with an obvious statement we need to show for $1\leq x\leq 2$ :

$$\frac{1}{3}f\left(x\right)+\frac{2}{3}f\left(\frac{3-x}{2}\right)\geq 1$$

In my first answer I show :

we have using a simple bound on logarithm ($1\leq x \leq 2$,$a=x,c=\frac{2}{x}$):

$$g(x)=\left((1-c)^{2}+ac(2-c)-ac(1-c)\cdot\frac{2\left(a-1\right)}{a+1}\right)^{2}\leq x^{\frac{4}{x}}$$

So it's not hard to show for $1\leq x\leq 2$ :

$$\frac{1}{3}g\left(x\right)+\frac{2}{3}j\left(\frac{\left(3-x\right)}{2}\right)\geq 1$$

Where as in my first answer we have for $x\in(0,1]$:

$$j(x)=x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)\leq f(x)\tag{K}$$ In expanding and using factorization .

To show $(K)$ we have for $x\in[0.8,1]$:

By Bernoulli's inequality :

$$x^{-5+5+\frac{4}{x}}\geq x^{5}\left(1+\left(\frac{4}{x}-5\right)\left(x-1\right)\right)\geq x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)$$

For the other interval we can use derivative and a bound for logarithm knowing that $j(x)$ can be negative.

It completes this detailed sketch of proof .

Reference :

Cirtoaje, V., Baiesu, A. An extension of Jensen's discrete inequality to half convex functions. J Inequal Appl 2011, 101 (2011). https://doi.org/10.1186/1029-242X-2011-101

Miss and Mister cassoulet char
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I show the inequality (almost algebraically speaking) for $x\in[0,4/3]$ :

$$4\left(x-1\right)+1\leq x^{\frac{4}{x}}$$

In fact we have a stronger inequality for $x\in[3/4,4/3]$:

$$\left(x^{-\frac{1}{4}}\left(1+\left(\frac{1}{x}+\frac{1}{4}\right)\left(x-1\right)\right)\right)^{4}\leq x^{\frac{4}{x}}\tag{I}$$

To show it $(I)$ we use Bernoulli's inequality simply remarking :

$$x^{\frac{4}{x}}=\left(x^{\frac{1}{x}+0.25-0.25}\right)^4$$

Now see WolframAlpha for a factorization .

Except using derivative I cannot show this sixth degree polynomials is positive for $x\in[3/4,4/3]$

In fact we have clearly for $1\le x\le 4/3$:

$$x^{6}+30x^{5}-687x^{4}+400x^{3}+1632x^{2}-1280x+256>x^{4}+30x^{4}-687x^{4}+400x^{2}+1632x^{2}-1280x+256$$

We can use the second derivative or substitute by $y=x^2$ and factorize again a part.

For $3/4\le x\leq 1$ the sixth degree polynomials is increasing via derivative .

Miss and Mister cassoulet char
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