$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$

Question

I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found? $$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$

My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.

Edited: I found some variants of the original problem.

Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.

Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.

Would it help to understand the function $f(x) = (x-1)(x-2)$? — Dhanvi Sreenivasan, Dec 18 '20 at 12:49
@DhanviSreenivasan My thoughts exactly. Note that equality is obtained when $a=b=c=1$. — Toby Mak, Dec 18 '20 at 12:52
I am not sure. Of course. The inequality could be written as $\sum (a -1)(a-2) \ge 0$, but I do not think we can continue with this. — andu eu, Dec 18 '20 at 12:53
@TobyMak Yes, but we can get that by setting $a = b = c$ and then we have $a ^ 3 = 1$, so $a = b = c = 1$. I am not entirely sure that that function could help. — andu eu, Dec 18 '20 at 13:03
It was just a thought, even I can't seem to get anything from it — Dhanvi Sreenivasan, Dec 18 '20 at 13:05
this is easily solvable by Buffalo Way or the $uvw$ method after homogenizing. Even SOS can solve it I am pretty sure. — dezdichado, Dec 18 '20 at 13:31
I can imagine all these variables ($a$, $b$, $c$) are $1$ at this point. The result would be $ 9 \ge 9 $ or $ 9 = 9 $. Does it specify that all variables have different values? — , Dec 18 '20 at 13:38

score 5 · Answer 1 · answered Dec 18 '20 at 14:25

Let $t = \sqrt{ab}$ and $$f(a,b,c) = a^2+b^2+c^2 + 6 - 3(a+b+c).$$ Suppose $c = \min\{a,\,b,\,c\},$ then $c \leqslant 1,$ so $$\sqrt{a}+\sqrt{b} \geqslant 2\sqrt{t} \geqslant 2.$$ We have $$f(a,b,c) - f(t,t,c) = a^2+b^2-2t^2 - 3(a+b-2t) = \left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\right]\left(\sqrt{a} - \sqrt{b}\right)^2 \geqslant 0.$$ Therefore $f(a,b,c) \geqslant f(t,t,c),$ and $$f(t,t,c) = f\left(t,t,\frac{1}{t^2}\right) = \frac{(2t^4-2t^3+2t+1)(t-1)^2}{t^4} \geqslant 0.$$ The proof is completed.

It is a nice solution. (+1) – River Li Dec 18 '20 at 15:30 — River Li, Dec 18 '20 at 15:30

River Li · Accepted Answer · 2020-12-18T14:03:20.723

3

By using Schur's inequality and the identity \begin{align} &(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca)\\ =\ & a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b), \end{align} we have $(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca) \ge 0$ which results in $$\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca. \tag{1}$$

We need to prove that $(a+b+c)^2 - 2(ab + bc + ca) + 6 \ge 3(a+b+c)$. By using (1), it suffices to prove that $$(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge 3(a+b+c)$$ that is (using $abc = 1$) $$\frac{(a+b+c - 3)[(a+b+c)^2 - 3(a+b+c) + 3]}{2(a+b+c)} \ge 0$$ which is true (using $a+b + c \ge 3\sqrt[3]{abc} = 3$).

We are done.

Remark: Actually, it is the pqr method.

edited Dec 18 '20 at 14:03

answered Dec 18 '20 at 13:53

River Li

37,323

1

Though I don't fully understand how this formula works, its pretty much convincing. Good job. – Dec 18 '20 at 13:57
Sorry, but I did not understand how proving a losened inequality does the proof. By, $(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge (a+b+c)^2 - 2(ab + bc + ca) + 6$, the inequality is lossening strength, could you explain something I am missing? – Book Of Flames Dec 18 '20 at 13:59
It is well-known in the pqr setting/method. From Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$ which is written as $p^3 - 4pq + 9r \ge 0$ where $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$. – River Li Dec 18 '20 at 14:00
I was actually saying that replacing $(a+b+c)^2 - 2(ab + bc + ca) + 6$ by $(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6$, the inequality is losing strength. That is, proving $(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge 3(a+b+c) $ is easier than proving $(a+b+c)^2 - 2(ab + bc + ca) + 6 \ge 3(a+b+c)$. – Book Of Flames Dec 18 '20 at 14:05
@BookOfFlames It should be $(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \le (a+b+c)^2 - 2(ab + bc + ca) + 6$ since $\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca$. – River Li Dec 18 '20 at 14:06
Oh, I see, Thanks! – Book Of Flames Dec 18 '20 at 14:07

score 1 · Answer 3 · answered Dec 19 '20 at 10:17

My second solution:

WLOG, assume $c = \min(a,b,c)$.

If $c \le \frac{1}{2}$, we have $$a^2+b^2+c^2 + 6 - 3(a+b+c) = (a-\tfrac{3}{2})^2 + (b - \tfrac{3}{2})^2 + c^2 - 3c + \tfrac{3}{2} \ge 0.$$

If $c > \frac{1}{2}$, noting that $x^2 + 2 - 3x + \ln x \ge 0$ for all $x > \frac{1}{2}$ (see the remark at the end), we have $$a^2+b^2+c^2 + 6 - 3(a+b+c) = \sum_{\mathrm{cyc}} (a^2 + 2 - 3a + \ln a) \ge 0.$$

We are done.

Remark: Let $f(x) = x^2 + 2 - 3x + \ln x$. We have $f'(x) = \frac{(2x-1)(x-1)}{x}$. Thus, $f(x)$ is strictly decreasing on $(\frac{1}{2}, 1)$, and strictly increasing on $(1, \infty)$. Also $f(1) = 0$. Thus, we have $f(x) \ge 0$ for all $x > \frac{1}{2}$.

score 0 · Answer 4 · answered Dec 19 '20 at 12:09

My third solution:

Actually $a^2 + b^2 + c^2 + 6 - 3(a+b+c) + (abc - 1) \ge 0$ for all $a, b, c \ge 0$.

(pqr method)

Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.

By Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$ which is written as $p^3 - 4pq + 9r \ge 0$, we have $\frac{p^3 + 9r}{4p}\ge q$.

We need to prove that $p^2 - 2q + 6 - 3p + r - 1 \ge 0$.

It suffices to prove that $p^2 - 2 \cdot \frac{p^3 + 9r}{4p} + 6 - 3p + r - 1 \ge 0$ or $$\frac{(2p-9)r}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0. \tag{1}$$

If $2p- 9 \ge 0$, clearly (1) is true.

If $2p - 9 < 0$, since $\frac{p^3}{27} \ge r$, it suffices to prove that $$\frac{(2p-9)\frac{p^3}{27}}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0$$ that is $$\frac{1}{27}(p+15)(p-3)^2 \ge 0.$$

We are done.

$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$

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