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\(\Leftrightarrow6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+20=\dfrac{5\left(x+y\right)\left(xy+3\right)}{xy}\ge\dfrac{5\left(x+y\right)2\sqrt{3xy}}{xy}=10\sqrt{3}\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right)\)
Đặt \(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=t\ge2\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\)
\(\Rightarrow6\left(t^2-2\right)+20\ge10\sqrt{3}t\)
\(\Rightarrow3t^2-5\sqrt{3}t+4\ge0\)
\(\Rightarrow\left(\sqrt{3}t-1\right)\left(\sqrt{3}t-4\right)\ge0\)
Do \(t\ge2\Rightarrow\sqrt{3}t-1>0\)
\(\Rightarrow\sqrt{3}t-4\ge0\Rightarrow t\ge\dfrac{4}{\sqrt{3}}\)
\(\Rightarrow t^2\ge\dfrac{16}{3}\Rightarrow t^2-2\ge\dfrac{10}{3}\)
\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge\dfrac{10}{3}\) (do \(\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\))
Vậy \(A_{min}=\dfrac{10}{3}\) khi \(\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
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\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
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\(2x^3+2y^3=x^3+x^5+y^3+y^5\ge2x^4+2y^4\)
\(\Rightarrow x^3+y^3\ge x^4+y^4\Rightarrow x^2+y^2+x^3+y^3\ge x^4+x^2+y^4+y^2\ge2x^3+2y^3\)
\(\Rightarrow x^2+y^2\ge x^3+y^3\Rightarrow x+y+x^2+y^2\ge x+x^3+y+y^3\ge2x^2+2y^2\)
\(\Rightarrow x+y\ge x^2+y^2\)
\(\Rightarrow x+y\ge x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\Rightarrow x^2-xy+y^2\le1\Rightarrow x^2+y^2\le1+xy\)
Dấu "=" xảy ra khi \(x=y=1\)
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Áp dụng BĐT AM-GM ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\)
Suy ra: \(P=6\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+8\left[\left(x^2+y^2\right)^2-2\left(xy\right)^2\right]+\frac{5}{xy}\)
\(\ge6\left(1-\frac{3}{4}\right)+8\left(\frac{1}{4}-\frac{1}{8}\right)+\frac{5}{\frac{1}{4}}\) (Do x+y=1) \(\Rightarrow P\ge6-\frac{9}{2}+2-1+20=\frac{45}{2}\)(đpcm).
Dấu "=" xảy ra <=> x=y=1/2.
ai giúp câu này với