Let $x,y,z\in \mathbb{R}^+$ be such that $xy+yz+zx=x+y+z$. Prove the following inequality: $\frac{1}{x^2+y+1}+\frac{1}{y^2+z+1}+\frac{1}{z^2+x+1}\leq 1$.
Problem
Source: VIII International Festival of Young Mathematicians Sozopol 2017, Theme for 10-12 grade
Tags: algebra, inequalities
bel.jad5
21.08.2019 10:43
Pinko wrote: Let $x,y,z\in \mathbb{R}^+$ be such that $xy+yz+zx=x+y+z$. Prove the following inequality: $\frac{1}{x^2+y+1}+\frac{1}{y^2+z+1}+\frac{1}{z^2+x+1}\leq 1$. According to Cauchy-Schwarz inequality, we get: $$\sum_{cyc} \frac{1}{x^2+y+1} \leq \sum_{cyc} \frac{1+y+z^2}{(x+y+z)^2}=\frac{3+(x+y+z)+(x^2+y^2+z^2)}{(x+y+z)^2} \leq \frac{2(xy+yz+zx)+(x^2+y^2+z^2)}{(x+y+z)^2}=1$$Equality at $x=y=z=1$.
sqing
24.08.2019 15:52
Let $ a,b$ be positive real numbers such that $ ab=1.$ Prove that$$\frac{2}{a+b+1}\geq\frac{1}{a^2+b+1}+\frac{1}{a+b^2+1}\geq\frac{2}{a^2+b^2+1}$$Let $ a,b,c$ be positive real numbers such that $abc=1.$ Prove that
Let $ x,y,z$ be positive real numbers such that $ xyz=1.$ Prove that
$$ \frac1{x^2 + y + 1} + \frac1{y^2 + z + 1} + \frac1{z^2 + x + 1}\ge\frac4{x +y+ z + 1}$$https://artofproblemsolving.com/community/c6h271832p13283388
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bel.jad5 wrote:
Pinko wrote:
Let $x,y,z\in \mathbb{R}^+$ be such that $xy+yz+zx=x+y+z$. Prove the following inequality:
$\frac{1}{x^2+y+1}+\frac{1}{y^2+z+1}+\frac{1}{z^2+x+1}\leq 1$.
According to Cauchy-Schwarz inequality, we get:
$$\sum_{cyc} \frac{1}{x^2+y+1} \leq \sum_{cyc} \frac{1+y+z^2}{(x+y+z)^2}=\frac{3+(x+y+z)+(x^2+y^2+z^2)}{(x+y+z)^2} \leq \frac{2(xy+yz+zx)+(x^2+y^2+z^2)}{(x+y+z)^2}=1$$Equality at $x=y=z=1$.
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