Vì x>0; y>0

Nên áp dụng BĐT Cô-si ta có: \(x+y\ge2\sqrt{xy}\)

\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{x}.\frac{1}{y}}=2\sqrt{\frac{1}{xy}}\)

Mà \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Nên \(\frac{1}{2}\ge2.\frac{1}{\sqrt{xy}}\Rightarrow\frac{1}{4}\ge\frac{1}{\sqrt{xy}}\)

\(\Rightarrow4\le\sqrt{xy}\) (C)

Ta có: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}\)

Thế (C) vào ta được: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{4}=4\)

Dấu "=" xảy ra <=> x = y

Vậy A_Min = 4 khi và chỉ khi x = y

\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow\frac{1}{2}>=\frac{4}{x+y}\Rightarrow x+y>=8\left(1\right)\)(bđt svacxo)

\(\frac{1}{x}+\frac{1}{y}>=2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}=\frac{2}{\sqrt{xy}}\Rightarrow\frac{1}{2}>=\frac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}>=4\Rightarrow2\sqrt{xy}>=8\left(2\right)\)(bđt cosi)

từ \(\left(1\right);\left(2\right)\Rightarrow x+2\sqrt{xy}+y>=8+8=16\Rightarrow\left(\sqrt{x}+\sqrt{y}\right)^2>=16\)

mà \(\sqrt{x}>0;\sqrt{y}>0\Rightarrow\sqrt{x}+\sqrt{y}>=4\)

dấu = xảy ra khi x=y=4

vậy min A là 4 khi x=y=4

A= \(\frac{x+2.3\sqrt{x}+9-11-2.3\sqrt{x}}{\sqrt{x}+3}\) =\(\sqrt{x}+3\)-\(\frac{11+6\sqrt{x}}{\sqrt{x}+3}\)=\(\sqrt{x}+3\)- \(\frac{6\sqrt{x}+18-7}{\sqrt{x}+3}\)

=\(\sqrt{x}+3\)-6 +\(\frac{7}{\sqrt{x}+3}\)= \(\sqrt{x}\)-3 +\(\frac{7}{\sqrt{x}+3}\)

A nguyen khi x nguyen duong  va 7 chia het \(\sqrt{x}+3\)

TH1: \(\sqrt{x}+3\)=1 (loai)

TH2: \(\sqrt{x}+3\)=7 =>x = 16

Theo giả thiết: \(a+b+c=3\Rightarrow b+c=3-a\). Tương tự: a+b=3-a và c+a=3-b

Khi đó \(\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}=\frac{1}{a^2-a+3}+\frac{1}{b^2-b+3}+\frac{1}{c^2-c+3}\)

Ta chứng minh BĐT phụ sau:

\(\frac{1}{a^2-a+3}\le\frac{4-a}{9}\)(1)

Thật vậy, BĐT (1) \(\Leftrightarrow9\le\left(4-a\right)\left(a^2-a+3\right)\)

\(\Leftrightarrow9\le-a^3+5a^2-7a+12\)\(\Leftrightarrow-a^3+5a^2-7a+3\ge0\)

\(\Leftrightarrow-a^3+a^2+4a^2-4a-3a+3\ge0\)

\(\Leftrightarrow-a^2\left(a-1\right)+4a\left(a-1\right)-3\left(a-1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(-a^2+4a-3\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(-a^2+a+3a-3\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)\left[-a\left(a-1\right)+3\left(a-1\right)\right]\ge0\)

\(\Leftrightarrow\left(a-1\right)^2\left(3-a\right)\ge0\)(2)

Ta thấy \(a;b;c>0\) và \(a+b+c=3\Rightarrow a< 3\)\(\Rightarrow3-a>0\)

Mà \(\left(a-1\right)^2\ge0\forall a\). Nên \(\left(a-1\right)^2\left(3-a\right)\ge0\)

Do đó: BĐT (2) luôn đúng với mọi 0<a<3 => BĐT (1) cũng đúng

Chứng minh tương tự \(\frac{1}{b^2-b+3}\le\frac{4-b}{9};\frac{1}{c^2-c+3}\le\frac{4-c}{9}\)

Từ đó suy ra:

\(\frac{1}{a^2-a+3}+\frac{1}{b^2-b+3}+\frac{1}{c^2-c+3}\le\frac{12-\left(a+b+c\right)}{9}=\frac{12-3}{9}=1\)(Do a+b+c=3)

=> ĐPCM.

Cho x,y,z € Z+ tm: x+y+z=4

Tính A= \(\sqrt{ }\)x(4-y)(4-z) +\(\sqrt{ }\)y(4-x)(4-x) +\(\sqrt{ }\)z(4-x)(4-y) -\(\sqrt{ }\)xyz

Nhân tử và mẫu của biểu thức với \(\sqrt{m}+\sqrt{n}-\sqrt{m+n}.\)

\(\Rightarrow\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}+\sqrt{m+n}\right)\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+n+2\sqrt{mn}-m-n}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)

Ta có: \(\frac{2\sqrt{mn}}{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)

\(=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+2\sqrt{mn}+n-m-n}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{2\sqrt{mn}}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)( đpcm )

Áp dụng: Với \(m=2\)và \(n=5\)và \(mn=10\); \(m+n=7\)ta có:

\(\frac{2\sqrt{10}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=\sqrt{2}+\sqrt{5}-\sqrt{2+5}=\sqrt{2}+\sqrt{5}-\sqrt{7}\)