Suppose positive real numers $x,y,z,w$ satisfy $(x^3+y^3)^4=z^3+w^3$. Prove that $$x^4z+y^4w\geq zw.$$
Problem
Source: 2021 Peru TST D3P1
Tags: inequalities, algebra
starchan
05.06.2022 21:24
We want to show that $\frac{x^4}{w}+\frac{y^4}{z} \geqslant 1$. Call the previous expression $E$; we have to prove $E \geqslant 1$ henceforth. Now we use Hölder's inequality to obtain that $(w^3+z^3) \cdot E^3 \geqslant (x^3+y^3)^4$. The $E \geqslant 1$ simply follows now.
sqing
02.12.2022 15:29
YII.I. wrote: Suppose positive real numers $x,y,z,w$ satisfy $(x^3+y^3)^4=z^3+w^3$. Prove that $$x^4z+y^4w\geq zw.$$
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a22886
02.12.2022 15:54
Holder does the trick
Manteca
27.10.2024 07:51
By Holder's Inequality we have that: $$(\frac{x^4}{w} + \frac{y^4}{z})^{\frac{3}{4}} (w^3 + z^3)^{\frac{1}{4}} \geq x^3 + y^3 \Rightarrow \frac{x^4}{w} + \frac{y^4}{z} \geq 1$$