$\text{Cho x,y,z là các số thực thỏa mãn x ≥ z CMR:}$ $\frac{xz}{y²+yz}$ + $\frac{y²}{xz+yz}$ + $\frac{x+2z}{x+z}$ $\geq$ $\frac{5}{2}$ *Copy=report,

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Ta cần chứng minh:}$

$\text{$\frac{xz}{y²+yz}$ + $\frac{y²}{xz+yz}$ + $\frac{z}{x+2}$ $\geq$ $\frac{3}{2}$}$

$\text{⇔ $\frac{1}{\frac{y}{x}+\frac{y}{z}+\frac{y}{x}}$ + $\frac{1}{\frac{x}{y}+\frac{z}{y}}$+$\frac{1}{\frac{x}{z}+1}$ $\geq$ $\frac{3}{2}$ (1)}$

$\text{Đặt $\frac{x}{y}$ = a , $\frac{y}{z}$ = b , $\frac{z}{x}$ = c}$

$\text{ĐK: a,b,c>0 và c≤1 do x≥z , abc=1}$

$\text{Khi đó (1) ⇔ $\frac{1}{\frac{b}{a}+\frac{1}{a}}$ + $\frac{1}{\frac{a}{b}+\frac{1}{b}}$+$\frac{1}{\frac{1}{c}+1}$ ≥ $\frac{3}{2}$}$

$\text{⇔ $\frac{a}{b+1}$ + $\frac{b}{a+1}$ + $\frac{c}{c+1}$ ≥ $\frac{3}{2}$}$ 

$\text{Ta sẽ đi chứng minh:}$

$\text{$\frac{a}{b+1}$ + $\frac{b}{a+1}$ $\geq$ $\frac{2\sqrt{ab}}{\sqrt{ab}+1}$}$

$\text{⇔ $\frac{a²+b²+a+b}{ab+a+b+1}$ $\geq$ $\frac{2\sqrt{ab}}{\sqrt{ab}+1}$}$

$\text{⇔ (a≥+b²+a+b)($\sqrt{ab}$)≥(a+bb+ab+1).2$\sqrt{ab}$}$ 

$\text{⇔a$\sqrt{ab}$+b²$\sqrt{ab}$+a$\sqrt{ab}$+b$\sqrt{ab}$+a²+b²+a+b≥2$\sqrt{ab}$.ab}$

$\text{+2$\sqrt{ab}$ .a+2$\sqrt{ab}$ .b+2$\sqrt{ab}$}$

$\text{⇔ $\sqrt{ab}$.(a-b)²+($\sqrt{a}$-$\sqrt{b}$)²+(a²+b²-$\sqrt{ab}$ .a-$\sqrt{ab}$ .b)≥0 (2)}$

$\text{Ta có a+b≥2ab(Cosi)}$

$\text{⇔ (a+b)²≥2$\sqrt{ab}$(a+b)}$

$\text{Mà (a+b)²≤2(a²+b²)}$

$\text{⇒ a²+b²≥$\sqrt{ab}$(a+b)}$

$\text{⇒ (2) >0 Dấu “=” xảy ra ⇔ a=b}$

$\text{⇒ (1) ≥ $\frac{2\sqrt{ab}}{\sqrt{ab}+1}$ + $\frac{c}{c+1}$ = $\frac{2.\frac{1}{\sqrt{c}}}{\frac{1}{\sqrt{c}}+1}$ + $\frac{c}{c+1}$ = $\frac{2}{\sqrt{c}+1}$ + $\frac{c}{c+1}$}$

$\text{= $\frac{2c+2+c\sqrt{c}+c}{(\sqrt{c}+1)(c+1)}$ (3)}$

$\text{Ta sẽ chứng minh (3) ≥ $\frac{3}{2}$}$

$\text{⇔ 2(3c+2+$c\sqrt{c}$)≥3($c\sqrt{c}$+$\sqrt{c}$+c+1)}$

$\text{⇔ 1-$3\sqrt{c}$+3c-$c\sqrt{c}$ ≥0}$

$\text{⇔ (1-$\sqrt{c}$)³≥0 ( Luôn đúng khi c≤1)}$

$\text{⇒ ĐPCM}$

$\text{Dấu “=” xảy ra ⇔ a=b=c=1⇔x=y=z}$