How to prove $(1+x)^7>7^{7/3} x^4, x>0$

Question

In 2004, IIT JEE asked the proof of $$[(1+a)(1+b)(1+c)]^7> 7^7 a^4 b^4 c^4, a,b,c>0,$$ which can be done by noticing that $$F=(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc \implies F-1=a+b+c+ab+bc+ca+abc.$$ By AM-GM of 7 items, we prove that $$(F-1)^7 \ge 7^7 a^4 b^4 c^4 \implies F^7 >7^7 a^4 b^4 c^4.$$ Here comes a question: How to prove $(1+x)^7>7^{7/3} x^4,~ x>0,$ otherwise.

Answer 1

Let $f(x):=\frac{(1+x)^7}{x^4}$; then $$ f'(x)=\frac{7(1+x)^6}{x^4}-\frac{4(1+x)^7}{x^5}=\frac{(1+x)^6(3x-4)}{x^5}. \tag{1} $$ Since $f'(x)<0$ if $0<x<\frac{4}{3}$ and $f'(x)>0$ if $x>\frac{4}{3}$, it follows that $f$ has a minimum at $x=\frac{4}{3}$, i.e. $$ f(x)\geq f\left(\frac{4}{3}\right)=\frac{7^7}{3^34^4}\quad\text{if $x>0$.} \tag{2} $$ It remains to show that $\frac{7^7}{3^34^4}>7^{7/3}$. This can done with a calculator, or by noticing that $$ \frac{7^7}{3^34^4}=\frac{7}{3}\left(\frac{49}{48}\right)^27^2>2\cdot 7^2, \tag{3} $$ whereas $$ 7^{7/3}=7^{1/3}7^2<8^{1/3}7^2=2\cdot 7^2. \tag{4} $$ It follows that, for $x>0$, $$ \frac{(1+x)^7}{x^4}>2\cdot 7^2>7^{7/3}\implies (1+x)^7>7^{7/3}x^4.\quad\square $$

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Addendum

Here is another proof, using the weighted AM-GM inequality: if $x>0$, \begin{align} (1+x)^7-(1+7x+21x^2+x^7)&=35x^3+35x^4+21x^5+7x^6 \\ &\geq 98\left((x^3)^{35}(x^4)^{35}(x^5)^{21}(x^6)^{7}\right)^{1/98} \\ &=98 x^4. \tag{A1} \end{align} It follows from $(\text{A}1)$ and $(4)$ that, if $x>0$, $$ (1+x)^7 > 98x^4 > 7^{7/3}x^4. \tag{A2} $$

Answer 2

We can also prove your inequality by applying the following Bernoulli’s inequality :

$(1+y)^r\leqslant1+ry\;\;\,$ for any $\;y>-1\,$ and $\;0\leqslant r\leqslant1\,.\quad\color{blue}{(*)}$

Let $\,x\,$ be any positive real number ($\,x>0\,$) .
From $\,(*)\,,\,$ for $\,y=\dfrac{3x}4-1\,$ and $\,r=\dfrac47\,,\,$ it follows that
$\left(\dfrac{3x}4\right)^{\!\!\frac47}\leqslant\dfrac37\big(1+x\big)\;\;,$

$\dfrac{3^4}{4^4}x^4\leqslant\dfrac{3^7}{7^7}\big(1+x\big)^{\!7}\;\;,$

$7^{\frac73}x^4<\dfrac{7^7}{3^3\!\!\cdot\!4^4}x^4\leqslant\big(1+x\big)^{\!7}\,.$

Answer 3

Note that $\binom74=35 < 7^{7/3}\approx93.4$ so a proof with the Binomial theorem "only" will not be so clean. In any case, it can be done: for reference observe $7^{7/3} \le 49 \times 8^{1/3} =98 $, and $$(1+x)^7 = 1 + 7x + 21 x^2 + 35x^3 + 35 x^4 + 21x^5 + 7 x ^6 + x^7.$$

First we consider $0\le x \le 1$. In which case, to get a lower bound, we can replace $1,x,x^2$ with $x^3$ and drop terms higher order than $x^4$: $$ (1+x)^7 \ge (\underbrace{1+7+21+35}_{=64})x^3 + 35x^4 \overset\star> 98 x^4.$$ The inequality marked $\star$ is correct iff $x<64/63$, and in particular if $x\le1$.

Next we treat $x>1$. We will need to first prove the result for sufficiently large $x$ and then bootstrap towards $x=1$.

• $x\ge 9/4$: Using $x>1$, we can replace $x^6,x^7,x^8$ with $x^5$ for a lower bound, and drop terms lower order than $x^4$: $$(1+x)^7 \ge 35x^4 + (\underbrace{1+7+21}_{29})x^5 \overset{\heartsuit}> 98x^4,$$ and the inequality marked $\heartsuit$ holds iff $x > \frac{63}{29}$ and in particular if $x \ge \frac{63}{28} = \frac94 = 2.25$.
• $\frac32 \le x < \frac94 $: In this region, instead of dropping terms, we can take advantage of $\frac{4x}9 < 1$ like the $x<1$ case and bound $1 > \frac{4x}9 > \frac{4^2x^2}{9^2}> \frac{4^3x^3}{9^3} > \dots $ (Note that $4^4/9^4$ is very small so we still drop it for simplicity): $$ (1+x)^7 \ge \underbrace{\left( \frac{7\times 4^3}{9^3} + \frac{21\times 4^2}{9^2} + \frac{35\times4}9 + 35\right)}_{=c}x^4 + 29x^5 \overset{\#}>98x^4 $$ and the inequality marked $\#$ holds iff $x>\frac{98-c}{29} = \frac{31115}{21141}$ and in particular if $x>\frac{31711.5}{21141} = 3/2 = 1.5$.
• Finally, $1 < x < \frac32$: repeating the above argument with $3x/2<1$ instead of $9x/4<1$: $$ (1+x)^7 \ge \underbrace{\left( \frac{2^4}{3^4} + \frac{7\times 2^3}{3^3} + \frac{21\times 2^2}{3^2} + \frac{35\times2}3 + 35\right)}_{=k}x^4 + 29x^5 >98x^4 $$ where the last inequality holds because $\frac{98 - k}{29} = \frac{2273}{2349} < 1$.

There is also a way to use Gaetano's Taylor series idea for $x>1$: "simply" observe that $$(1+x)^7 - 7^{7/3} x^4\\ = (128 - 49 \sqrt[3]7) + (448 - 196 \sqrt[3]7) (x - 1) + (672 - 294 \sqrt[3]7) (x - 1)^2 + (560 - 196 \sqrt[3]7) (x - 1)^3 \\ + (280 - 49 \sqrt[3]7) (x - 1)^4 + 84 (x - 1)^5 + 14 (x - 1)^6 + (x - 1)^7 $$ As previously mentioned, $7^{1/3} < 2$, so each coefficient of the powers of $(x-1)$ is positive. Hence, $(1+x)^7 - 7^{7/3} x^4>0$ for $x>1$, QED.

Answer 4

Let consider the function:

$$\forall x>0,\Delta(x) := (1+x)^7-7^{7/3}x^4.$$

This function is differentiable. The derivatives are given by (you can use a software):

$$\begin{cases} \Delta^{(5)}(x) = 2520(x+1)^2\\ \Delta^{(4)}(x) = 24(35(x+1)^3-7^{1/3})\\ \Delta^{(3)}(x) = 6 \left(- 4 \cdot \sqrt[3]{7} x + 35 \left(x + 1\right)^{4}\right)\\ \Delta^{(2)}(x) = 6 \left(- 2 \cdot \sqrt[3]{7} x^{2} + 7 \left(x + 1\right)^{5}\right)\\ \Delta^{(1)}(x) = - 4 \cdot \sqrt[3]{7} x^{3} + 7 \left(x + 1\right)^{6}\\ \end{cases}$$

You can notice that $\Delta^{(5)}$ is postive. Then $\Delta^{(4)}$ is increasing, since $\Delta^{(4)}(0)$ is positive, $\Delta^{(4)}$ is then positive on $(0,\infty)$. By applying the same reasoning to all the derivatives (easy recursion), you have the desired result.

You could also argue that all the derivatives in 0 are positive, the Taylor expansion (finite sum since the function is polynomial) is then positive on $\mathbb{R} _-$.

This method is quite unelegant but is a proof anyway.