Question

Demonstrati ca, daca x,y,z∈R+, atunci:
a) x²+y²≥2xy
b) x+y≥2√xy
c) $\frac{xy}{x+y} \leq \frac{x+y}{4}$
d) x³+y³≥xy(x+y)

Answer 1

$\displaystyle\bf\\a)~x^2+y^2\geq 2xy,~x,y \in \mathbb{R}_+.\\x^2+y^2\geq 2xy \Leftrightarrow x^2+y^2-2xy\geq 0 \Leftrightarrow (x-y)^2\geq 0,~evident.\\b)~x+y \geq 2\sqrt{xy},~x,y\in \mathbb{R}_+.\\x+y \geq 2\sqrt{xy} \Leftrightarrow x+y - 2\sqrt{xy} \geq 0 \Leftrightarrow (\sqrt{x} -\sqrt{y} )^2 \geq 0,~evident.\\c)\frac{xy}{x+y} \leq \frac{x+y}{4},~x,y\in \mathbb{R}_+.\\\frac{xy}{x+y} \leq \frac{x+y}{4} \Leftrightarrow 4xy \leq (x+y)^2 \Leftrightarrow 4xy \leq x^2+y^2+2xy \Leftrightarrow$

$\displaystyle\bf\\0\leq x^2+y^2-2xy \Leftrightarrow 0\leq (x-y)^2,~evident.\\d) ~x^3+y^3 \geq xy(x+y),~x,y\in \mathbb{R}_+.\\x^3+y^3 \geq xy(x+y) \Leftrightarrow (x+y)(x^2-xy+y^2)\geq xy(x+y),~\Leftrightarrow\\\frac{(x+y)(x^2-xy+y^2)}{(x+y)} \geq xy \Leftrightarrow x^2-xy+y^2 \geq xy \Leftrightarrow x^2-2xy+y^2\geq 0 \Leftrightarrow\\(x-y)^2\geq 0,~evident.$