If $a,b,c$ are positive real numbers, then prove that $[(1+a)(1+b)(1+c)]^7$>$7^7a^4b^4c^4$. I am stuck on this question and do not know how to proceed. Any hint or clue how I might tackle this question
How to Prove that $[(1+a)(1+b)(1+c)]^7$>$7^7a^4b^4c^4$
-1
$\begingroup$
inequality
substitution
a.m.-g.m.-inequality
holder-inequality
2 Answers
2
Let $abc=x^{21}$.
Hence, by Holder $$\sum_{cyc}(1+a)\geq\left(1+\sqrt[3]{abc}\right)^3=(1+x^7)^3.$$ Thus, it remains to prove that $$1+x^7>\sqrt[3]7x^4,$$ which follows from AM-GM: $$1+x^7=3\cdot\frac{1}{3}+4\cdot\frac{x^7}{4}\geq7\sqrt[7]{\frac{x^{28}}{3^3\cdot4^4}}>\sqrt[3]7x^4.$$ Done!
2
We have to prove $$(1+a)(1+b)(1+c)>7 (abc)^{\frac {4}{7}} $$ $$\Rightarrow 1+\sum a +\sum ab + abc >7 (abc)^{\frac {4}{7}} $$ We know by AM-GM inequality $$\sum a \geq 3 (abc)^{\frac {1}{3}} \text { and } \sum ab \geq 3 (abc)^{\frac {2}{3}} $$ Hence, $$\text {LHS } \geq 1+ 3 (abc)^{\frac {1}{3}} +3 (abc)^{\frac {2}{3}} + abc = 1+ (abc)^{\frac {1}{3}} + (abc)^{\frac {1}{3}} + (abc)^{\frac {1}{3}} + (abc)^{\frac {2}{3}} + (abc)^{\frac {2}{3}} + (abc)^{\frac {2}{3}} + abc \geq 7\sqrt [7]{(abc)^2 (abc)(abc)} =7 (abc)^{\frac {4}{7}} $$ Hope it helps.