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Lời giải:
Ta có:
\(\frac{4x^2y^2}{(x^2+y^2)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\geq 3\)
\(\Leftrightarrow \frac{4x^2y^2}{(x^2+y^2)^2}-1+\frac{x^2}{y^2}+\frac{y^2}{x^2}-2\geq 0\)
\(\Leftrightarrow \frac{4x^2y^2-(x^2+y^2)^2}{(x^2+y^2)^2}+\left(\frac{x}{y}-\frac{y}{x}\right)^2\geq 0\)
\(\Leftrightarrow \frac{-(x^2-y^2)^2}{(x^2+y^2)^2}+\frac{(x^2-y^2)^2}{x^2y^2}\geq 0\)
\(\Leftrightarrow (x^2-y^2)^2\left(\frac{1}{x^2y^2}-\frac{1}{(x^2+y^2)^2}\right)\geq 0\)
\(\Leftrightarrow \frac{(x^2-y^2)^2(x^4+y^4+x^2y^2)}{x^2y^2(x^2+y^2)^2}\geq 0\) (luôn đúng)
Do đó ta có đpcm.
Dấu bằng xảy ra khi $x=y$
\(A=\dfrac{4x^2y^2}{\left(x^2+y^2\right)^2}+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
x,y khác 0
<=>\(A=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}\right)^2+\left(\dfrac{y}{x}\right)^2\)
\(A+2=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=m\)
\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=t;t\ge4\)
\(m=\dfrac{4}{t}+t\Leftrightarrow t^2-mt+4=0\)
f(t) có nghiệm t>= 4<=>\(\left\{{}\begin{matrix}m^2-16\ge0\\\dfrac{m+\sqrt{m^2-16}}{2}\ge4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|m\right|\ge4\\m^2-16\ge m^2-16m+64\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left|m\right|\ge4\\m\ge5\end{matrix}\right.\) \(\Leftrightarrow A+2\ge5;A\ge3=>dpcm\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1\right)+\left(\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\right)\ge0\)\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\) (đúng)
cách khác
đặt \(\dfrac{x}{y}+\dfrac{y}{x}=t\Rightarrow\left|t\right|\ge2\)
\(\Leftrightarrow t^2-3t+2\ge0\)
\(\Leftrightarrow\left(t-1\right)\left(t-2\right)\ge0\)
điều này luôn đúng với mọi |t| >=2 => dpcm
kết luận điều kiện đề hơi thừa
cái cần c/m đúng với mọi x,y khác 0
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\right)\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge\dfrac{6x}{y}+\dfrac{6y}{x}\)
\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-4.\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0^{\left(1\right)}\)
\(^{\left(1\right)}\)đúng \(\Rightarrowđpcm\)
Áp dụng BĐT : x4 + y4 ≥ 2x2y2
=> \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) ≥ 2 ( x , y > 0 )
TT , \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2 ( x , y > 0 )
Ta có : \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) + 4 ≥ 6 ( 1 )
\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\) ≥ 6 ( 2 )
Từ ( 1 ; 2) => đpcm
\(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2}{x^2y^2}\ge\dfrac{3\left(x-y\right)^2}{xy}\)
\(\Leftrightarrow\dfrac{\left[\left(x-y\right)\left(x+y\right)\right]^2}{x^2y^2}-\dfrac{3\left(x-y\right)^2}{xy}\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{\left(x+y\right)^2}{x^2y^2}-\dfrac{3}{xy}\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{x^2+y^2-xy}{x^2y^2}\right)\ge0\)( luôn đúng )
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