CM: \(\left(x+\dfrac{2}{y}\right)\left(\dfrac{y}{x}+2\right)\ge8\) ; x, y>0
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Lời giải:
Áp dụng BĐT Cô-si với \(x; \frac{1}{x}\) là hai số dương:
\(x+\frac{1}{x}\geq 2\sqrt{x.\frac{1}{x}}=2\)
\(\Rightarrow \left(x+\frac{1}{x}\right)^2\geq 4\)
Tương tự, \(\left(y+\frac{1}{y}\right)^2\geq 4\)
\(\Rightarrow \left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq 8\) (đpcm)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x=\frac{1}{x}\\ y=\frac{1}{y}\end{matrix}\right.\Leftrightarrow x=y=1\)
P.s: Có thể thấy điều kiện $x+y=2$ là dư thừa.
Hem thừa .-.
\(\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\ge\dfrac{\left(x+y+\dfrac{4}{x+y}\right)^2}{2}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a)
Với \(x>1\Rightarrow x-1>0\). Áp dụng BĐT AM-GM:
\(x=(x-1)+1\geq 2\sqrt{x-1}\)
\(\Rightarrow \frac{\sqrt{x-1}}{x}\leq \frac{\sqrt{x-1}}{2\sqrt{x-1}}=\frac{1}{2}\) (đpcm)
Dấu bằng xảy ra ki \(x-1=1\Leftrightarrow x=2\)
b) Trước tiên, ta có bđt phụ sau:
\(x^3+y^3\geq xy(x+y)\)
\(\Leftrightarrow (x-y)^2(x+y)\geq 0\) (luôn đúng với mọi \(x,y>1\) )
Do đó, \(\frac{x^3+y^3-(x^2+y^2)}{(x-1)(y-1)}\geq \frac{xy(x+y)-x^2-y^2}{(x-1)(y-1)}\geq 8\)
\(\Leftrightarrow xy(x+y)-(x^2+y^2)\geq 8(x-1)(y-1)\)
\(\Leftrightarrow x^2(y-1)+y^2(x-1)-8(x-1)(y-1)\geq 0\)
\(\Leftrightarrow (y-1)[x^2-4(x-1)]+(x-1)[y^2-4(y-1)]\geq 0\)
\(\Leftrightarrow (y-1)(x-2)^2+(x-1)(y-2)^2\geq 0\)
(luôn đúng với mọi \(x,y>1\) )
Do đó ta có đpcm
Dấu bằng xảy ra khi \(x=y=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)
Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)
b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)
=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
\(\left(x+\dfrac{2}{y}\right)\left(\dfrac{y}{x}+2\right)\ge2\sqrt{\dfrac{2x}{y}}.2\sqrt{\dfrac{2y}{x}}=2.2.2=8\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)