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\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)
\(P=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{1+\dfrac{y}{x}}\)
Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)
\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)
Dấu "=" xảy ra khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)
Nguyễn Việt Lâm Giáo viên làm thế nào để có thể nghĩ được ra như vậy?
\(\Leftrightarrow2P=6x+4y+\dfrac{12}{x}+\dfrac{16}{y}\\ \Leftrightarrow2P=\left(\dfrac{12}{x}+3x\right)+\left(\dfrac{16}{y}+y\right)+3\left(x+y\right)\\ \Leftrightarrow2P\ge2\sqrt{36}+2\sqrt{16}+3\cdot6=12+8+18=38\\ \Leftrightarrow P\ge19\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}3x^2=12\\y^2=16\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Điểm rơi: \(x=y=\frac{\sqrt{2}}{2}\)
Ta tách biểu thức được như sau: \(A=x+\frac{1}{x}+y+\frac{1}{y}=(x+\frac{1}{2x})+(y+\frac{1}{2y})+\frac{1}{2}(\frac{1}{2x}+\frac{1}{2y})\)
\(\geq 2\sqrt{x.\frac{1}{2x}}+2\sqrt{y.\frac{1}{2y}}+\frac{1}{2}.\frac{4}{x+y}=2\sqrt{2}+\frac{2}{x+y}\)
Áp dụng bất đẳng thức Bunhiacốpxki, ta lại có:
\((x+y)^2\leq 2(x^2+y^2)=2 \Rightarrow x+y\leq \sqrt{2}\)
\(\Rightarrow A\geq 3\sqrt{2}\)
Dấu bằng xảy ra khi \(x=y=\frac{\sqrt{2}}{2}\)
`<=>2P=10x+6y+24/x+32/y`
`<=>2P=6x+24/x+2y+32/y+4x+4y`
`<=>2P=6(x+4/x)+2(y+16/y)+4(x+y)`
Áp dụng BĐT cosi:
`x+4/x>=4=>6(x+4/x)>=24`
`y+16/y>=8=>2(y+16/y)>=16`
Mà `x+y>=6=>4(x+y)>=24`
`=>2P>=24+16+24=64`
`=>P>=32`
Dấu "=" `<=>x=2,y=4`
\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)
\(=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{\left(x+y\right)^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\left(1.\sqrt{x}+1.\sqrt{y}\right)}\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\sqrt{\left(1^2+1^2\right)\left(x+y\right)}}=\dfrac{1}{\dfrac{1}{2}\sqrt{2}}=\sqrt{2}\)
"=" khi x = y = 1/2
Áp dụng bất đẳng thức AM - GM:
\(P=4x+3y+\dfrac{6}{x}+\dfrac{9}{2y}\)
\(=\left(\dfrac{3}{2}x+\dfrac{6}{x}\right)+\left(\dfrac{1}{2}y+\dfrac{9}{2y}\right)+\left(\dfrac{5}{2}x+\dfrac{5}{2}y\right)\)
\(\ge2\sqrt{\dfrac{3}{2}x\times\dfrac{6}{x}}+2\sqrt{\dfrac{1}{2}y\times\dfrac{9}{2y}}+\dfrac{5}{2}\times5\)
\(=\dfrac{43}{2}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3}{2}x=\dfrac{6}{x}\\\dfrac{1}{2}y=\dfrac{9}{2y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\left(\text{nhận}\right)\)
Vậy \(Min_P=\dfrac{43}{2}\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Lời giải:
Ta có: $A=x^2+\frac{1}{y(x-y)}$. Đặt $x-y=a$ với $a>0$ thì áp dụng BĐT AM-GM ta có:
$A=(a+y)^2+\frac{1}{ay}\geq 4ay+\frac{1}{ay}\geq 2\sqrt{4ay.\frac{1}{ay}}=4$
Vậy $A_{\min}=4$ khi $x=\sqrt{2}; y=\frac{1}{\sqrt{2}}$
\(25P=\dfrac{x\left(2+3\right)^2}{2x+x+y+z}+\dfrac{y\left(2+3\right)^2}{2y+x+y+z}+\dfrac{z\left(2+3\right)^2}{2z+x+y+z}\)
\(25P\le x\left(\dfrac{2^2}{2x}+\dfrac{3^2}{x+y+z}\right)+y\left(\dfrac{2^2}{2y}+\dfrac{3^2}{x+y+z}\right)+z\left(\dfrac{2^2}{2z}+\dfrac{3^2}{x+y+z}\right)\)
\(25P\le6+\dfrac{9\left(x+y+z\right)}{x+y+z}=15\)
\(\Rightarrow P\le\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(x=y=z\)
\(P=3\left(x+\dfrac{9}{x}\right)+\left(y+\dfrac{16}{y}\right)+\left(x+y\right)\)
\(P\ge3.2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{y}}+7=33\)
\(P_{min}=33\) khi \(\left(x;y\right)=\left(3;4\right)\)