Áp dụng BĐT AM-GM:

\(P=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\)

\(=\dfrac{x^2}{y-1}+4\left(y-1\right)+\dfrac{y^2}{x-1}+4\left(x-1\right)-4\left(x+y\right)+8\)

\(\ge2\sqrt{\dfrac{x^2}{y-1}.4\left(y-1\right)}+2\sqrt{\dfrac{y^2}{x-1}.4\left(x-1\right)}-4\left(x+y\right)+8\)

\(\ge4\left(x+y\right)-4\left(x+y\right)+8=8\)

\(\Rightarrow P_{min}=8\Leftrightarrow x=y=2\)

\(\dfrac{x^2}{y-1}+4\left(y-1\right)\ge4x\) ; \(\dfrac{y^2}{x-1}+4\left(x-1\right)\ge4y\)

Cộng vế:

\(P+4\left(x+y\right)-8\ge4\left(x+y\right)\Rightarrow P\ge8\)

Dấu "=" xảy ra khi \(x=y=2\)

Áp dụng cosi

`1/x^2+1/y^2>=2/(xy)`

`=>1/2>=2/(xy)`

`=>xy>=4`

Aps dụng cosi

`=>x+y>=2\sqrt{xy}=2.2=4`

Dấu "=" xảy ra khi `x=y=4`

Có : \(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2\sqrt{\dfrac{1}{x^2}\cdot\dfrac{1}{y^2}}=\dfrac{2}{xy}\)

\(\Rightarrow xy\ge4\)

Ta có : \(A=x+y\ge2\sqrt{xy}=2\sqrt{4}=4\)

Dấu "=" xảy ra khi \(x=y=2\)

Vậy min A = 4 khi $x=y=2$

\(P=\left[\left(\dfrac{-1}{3}\right)^2x^3+\left(2x^2\right)^2+\dfrac{1}{2}\right]-\left[x\left(\dfrac{1}{3}x\right)^2+\dfrac{3}{2^3}+x^4\right]+\left(y-2013\right)^2=\left(\dfrac{1}{9}x^3+4x^4+\dfrac{1}{2}\right)-\left(\dfrac{1}{9}x^3+x^4+\dfrac{3}{8}\right)+\left(y-2013\right)^2=3x^4+\dfrac{1}{8}+\left(y-2013\right)^2\ge\dfrac{1}{8}\).

Dấu "=" xảy ra khi x = 0; y = 2013.

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\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

A>=1/(1+xy)=1/2

Dấu = xảy ra khi x=y=1

\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)

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

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

"=" khi x = y = 1/2

giúp mình voi ah

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

\(K\ge2\sqrt{\dfrac{4xy}{4xy}}+\dfrac{4}{x^2+y^2+2xy}+\dfrac{5}{\left(x+y\right)^2}\ge2+4+5=11\)

\(K_{min}=11\) khi \(x=y=\dfrac{1}{2}\)

\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)