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Áp dụng BĐT Cauchy-Schwarz,ta có:\(\left(\frac{x^2}{a}+\frac{y^2}{b}\right)\left(a+b\right)\ge\left(x+y\right)^2\). Chia hai vế cho a, b.Ta được:

\(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}^{\left(đpcm\right)}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{a}=\frac{y}{b}\)

Mấy cái dấu "=" anh tự xét.

Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)

a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)

b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)

gọi A là VT

Ta có : \(A=\left[\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\right]+\left[\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\right]-1\)

Áp dụng BĐT Cô-si,ta có :

\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\Rightarrow\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\ge0\)

\(\frac{x^{16}+y^{16}}{4}\ge\frac{x^8y^8}{2}=\left(\frac{x^8y^8}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)-\frac{3}{2}\ge4\sqrt[4]{\frac{x^8y^8}{16}}-\frac{3}{2}==2x^2y^2-\frac{3}{2}\)

\(\Rightarrow\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\ge\frac{-3}{2}\)

Từ đó ta có : \(A\ge0-\frac{3}{2}-1=\frac{-5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y\\x^2y^2=1\end{cases}\Leftrightarrow x=y=\pm1}\)

 dễ cm bđt: x²+y² ≥ (x+y)²/2, khai triễn là ra hằng đẳng đúng, dấu "=" khi x = y 
ad: P = (x+1/x)² + (y+1/y)² ≥ [x+1/x + y+1/y]²/2 = [(x+y) + (x+y)/xy]²/2 (*) 
bđt côsi: 1 = x+y ≥ 2√(xy) => 1 ≥ 4xy => 1/xy ≥ 4 
thay vào (*): P ≥ [1 + 1/xy]²/2 ≥ [1 + 4]²/2 = 25/2 (đpcm), dấu "=" khi x = y = 1/2 

Đặt \(P=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

Áp dụng bđt bunhiacopxki ta có:

\(\left[\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\right]\left(1^2+1^2\right)\ge\left[\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)\right]^2\)

\(\Leftrightarrow2P\ge\left(1+\frac{1}{x}+\frac{1}{y}\right)^2\)(1)

Ta có BĐT:\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\)( bạn tự CM = cách chuyển vế nhé )

Áp dụng bđt cô si cho 2 số dương x,y ta có:
\(\sqrt{xy}\le\frac{x+y}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}\ge4\)(2)

Thay (2) vào (1) ta được:

\(2P\ge25\)

\(\Rightarrow P\ge\frac{25}{2}\left(đpcm\right)\)