Let $x, y, z$ be nonnegative real numbers such that $x + y + z = 3$. Prove that $$\frac{x}{4-y}+\frac{y}{4-z}+\frac{z}{4-x}+\frac{1}{16}(1-x)^2(1-y)^2(1-z)^2\leq 1,$$and determine all such triples $(x, y, z)$ where the equality holds.
Problem
Source: 2020 Thailand TST 1.5
Tags: inequalities, algebra
sqing
09.03.2022 16:18
Let $x, y, z$ be nonnegative real numbers such that $x + y + z = 3$. Prove that $$\frac{x}{4-y}+\frac{y}{4-z}+\frac{z}{4-x}+\frac{1}{15}(1-x)^2(1-y)^2(1-z)^2\leq 1 $$$$\frac{x}{4-y}+\frac{y}{4-z}+\frac{z}{4-x}+(1-x)^2(1-y)^2(1-z)^2\leq \frac{19}{4}$$
Quidditch wrote:
Let $x, y, z$ be nonnegative real numbers such that $x + y + z = 3$. Prove that $$\frac{x}{4-y}+\frac{y}{4-z}+\frac{z}{4-x}+\frac{1}{16}(1-x)^2(1-y)^2(1-z)^2\leq 1,$$and determine all such triples $(x, y, z)$ where the equality holds.
sqing
09.03.2022 16:57
Let $x, y$ be nonnegative real numbers such that $x + y= 2$. Prove that $$ \frac{x}{3-y}+\frac{y}{3-x}+ \frac{1}{3}(1-x)^2(1-y)^2\leq1$$$$ \frac{x}{3-y}+\frac{y}{3-x}+(1-x)^2(1-y)^2\leq \frac{5}{3}$$
sqing
19.03.2022 14:53
sqing wrote: Let $x, y$ be nonnegative real numbers such that $x + y= 2$. Prove that $$ \frac{x}{3-y}+\frac{y}{3-x}+ \frac{1}{3}(1-x)^2(1-y)^2\leq1$$
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