Áp dụng BĐT Cô si cho 2 số dương a,b ta có \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(\dfrac{1}{a}+\dfrac{1}{b}\ge2.\sqrt{\dfrac{1}{a}.\dfrac{1}{b}}=>\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\dfrac{1}{a}.\dfrac{1}{b}}\)

suy ra \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\).Áp dụng vào bài toán ta có :\(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge\dfrac{4}{x^2+xy+y^2+xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (Do \(x+y\le1\))

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge\dfrac{\left(1+1\right)^2}{x^2+2xy+y^2}=\dfrac{4}{\left(x+y\right)^2}\ge\dfrac{4}{1}=4\)

\(VT=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{4039}{2xy}\)

\(VT\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{4039}{2.\dfrac{1}{4}\left(x+y\right)^2}=\dfrac{8082}{\left(x+y\right)^2}\ge\dfrac{8082}{1^2}=8082\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)

\(\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{5}{4.\dfrac{\left(x+y\right)^2}{4}}\)

\(\ge\dfrac{4}{1^2}+2+\dfrac{5}{1^2}\) (do \(x+y\le1\))

\(=11\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy GTNN của A là 11.

Đặt : A = 1/x^2+xy + 1/y^2+xy

Có : A = 1/x.(x+y) + 1/y.(x+y) = 1/x + 1/y ( vì x+y = 1 )

Áp dụng bđt 1/a + 1/b >= 4/a+b với mọi a,b > 0 cho x,y > 0 thì :

A >= 4/x+y = 4/1 = 4

Dấu "=" xảy ra <=> x=y=1/2

=> ĐPCM

Tk mk nha

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{2xy}\ge\dfrac{4}{1^2}+\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}\ge4+2=6\)

Dấu "=" xảy ra <=> x = y = 0,5

Áp dụng bđt : Với a>0 ; b>0 thì 1/b + 1/b >=4/(a+b) ta có :

\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{x^2+xy+y^2+xy}=\frac{4}{\left(x+y\right)^2}\ge4\)( vì 0 = < x + y <=1)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)

không biết :))))