Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.

Problem. Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.

I don’t know how to solve this problem... What I can think of is to just simplify this inequality: $$ \left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}. $$ How can I proceed with solving this problem?

Note: This is a question of sequence and series, specifically AM-GM-HM inequality...

Solution 1:

Notice $$\frac{(1+a)(1+b)(1+c)-1}{7}=\frac{a+b+c+ab+bc+ca+abc}{7} \ge \sqrt[7]{a^4b^4c^4}$$ Because of $\text{AM-GM}$.

We conclude

$$\frac{(1+a)(1+b)(1+c)}{7}>\frac{(1+a)(1+b)(1+c)-1}{7} \ge\sqrt[7]{a^4b^4c^4}$$

This is equivalent to

$$ \left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}. $$

Solution 2:

Consider the function $f(a)=\frac{(1+a)^7}{a^4}$.

We must prove that $f(a)\geq 7^{7/3}$ for all $a>0$.

to minimize this function we derive.

$f'(a)=\frac{(a+1)^6(3a-4)}{a^5}$.

So the global minimum for $f$ in $(0,\infty)$ is $f(\frac{4}{3})=\frac{(7/3)^7}{(4/3)^4}=\frac{7^7}{3^34^4}$

So we just have to prove $3^34^4\leq 7^{14/3}$:

Notice $3^34^4=27\times 256=6912$

Notice $7^{14/3}= 7^{4}\times 7^{2/3}\geq2401\times 3=7203$.

(To see $7^{2/3}\geq 3$ notice $3^3\leq 49$)