Question

Cho x,y > 0. Chứng minh rằng \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\geq 3\left(\frac{x}{y}+\frac{y}{x}\right)\)

Answer 1

Đặt \(\dfrac{x}{y}+\dfrac{y}{x}=a\)\(\Rightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=a^2\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}=a^2-2\)

Ta có \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4=a^2-2+4=a^2+2\)

\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)=3a\)

Ta có \(a^2+2-3a=a^2-2.a.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}=\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)

lạ có \(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2}{xy}-\dfrac{2xy}{xy}+\dfrac{y^2}{xy}=\dfrac{\left(x-y\right)^2}{xy}\ge0\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge2\)\(\Rightarrow a\ge2\Rightarrow a-\dfrac{3}{2}\ge\dfrac{1}{2}\)\(\Rightarrow\left(a-\dfrac{3}{2}\right)^2\ge\dfrac{1}{4}\Rightarrow\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge0\)

\(\Rightarrow a^2+2-3a\ge0\Rightarrow a^2+2\ge3a\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

Answer 2

\(\left\{{}\begin{matrix}x;y>0\\\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)

từ (2) có \(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2.\dfrac{x}{y}.\dfrac{y}{x}+\dfrac{y^2}{x^2}\right)+2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\right]-\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)-2\right]\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)\ge0\) (3)

từ (1) có \(\dfrac{x}{y}+\dfrac{y}{x}=\left(\sqrt{\dfrac{x}{y}}-\sqrt{\dfrac{y}{x}}\right)^2+2\ge2\) (4)

từ (4) ; \(\left\{{}\begin{matrix}\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)>0\\\dfrac{x}{y}+\dfrac{y}{x}-2\ge0\end{matrix}\right.\) (I)

từ (I) => (3) đúng mọi phép biến đổi là <=> đẳng thức khi \(\dfrac{x}{y}=\dfrac{y}{x}\Rightarrow x=y\)=> dpcm

Answer 3

ta có: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+2=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2\)(1)

vì x,y >0

nên \(\dfrac{x}{y}+\dfrac{y}{x}=t,t\ge2\)ta được:

\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2+2\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

<=>\(t^2+2\ge3t< =>t^2+2-3t\ge0< =>t^2-t-2t+2\ge0< =>t\left(t-1\right)-2\left(t-1\right)\ge0< =>\left(t-1\right)\left(t-2\right)\ge0\)

(BĐT cuối luôn đúng vì t > hoặc = 0) (2)

từ (1) và (2)=>\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)