Ta có : \(VP=\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{yx}}=2\)

Vậy \(Q_{min}=2\)với \(x=y\)

mình không chắc về phân bđt này lắm

Đặt x=a, \(\frac{1}{y}=b\)\(\Rightarrow a+b\le1\)

Ta có: \(Q=ab+\frac{1}{ab}=16ab+\frac{1}{ab}-15ab\ge2\sqrt{\frac{16ab}{ab}}-\frac{15.\left(a+b\right)^2}{4}=8-\frac{15.1}{4}=\frac{17}{4}\)

Dấu "=" xảy ra khi a=b=\(\frac{1}{2}\)hay \(x=\frac{1}{2},y=2\)

fevg7yghrudhvgryhde7777777777777777777777777777777777777777777777777777444444444444444444444444444444444444444444yyyyyyyyyyyyyyyyyyyyyyyy7555555555555555555

ap dung bdt cauchy schwarz ta co

\(\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}>=\frac{\left(x-1+z-1+y-1\right)^2}{x+y+z}=\frac{1}{2}\)

vay min=1/2

\(Q\ge2xy+\frac{2}{xy}=2xy+\frac{1}{8xy}+\frac{15}{8xy}\ge2\sqrt{\frac{2xy}{8xy}}+\frac{15}{2\left(x+y\right)^2}\ge1+\frac{15}{2}=\frac{17}{2}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

By Titu's Lemma we easy have:

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

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

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

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

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

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

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

\(1\ge\frac{1}{x}+\frac{1}{y+1}\ge\frac{4}{x+y+1}\Rightarrow x+y+1\ge4\)

\(\Rightarrow x+y\ge3\)

\(P=\frac{x+y}{9}+\frac{1}{x+y}+\frac{8}{9}\left(x+y\right)\ge2\sqrt{\frac{x+y}{9\left(x+y\right)}}+\frac{8}{9}.3=\frac{10}{3}\)

\(P_{min}=\frac{10}{3}\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(2\sqrt{xy}\le x+y\le1\Rightarrow\frac{1}{\sqrt{xy}}\ge2\Rightarrow\frac{1}{xy}\ge4\)

\(P\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}=2\sqrt{\frac{15}{16xy}+\frac{1}{16xy}+xy}\)

\(P\ge2\sqrt{\frac{15}{16}.4+2\sqrt{\frac{xy}{16xy}}}=\sqrt{17}\)

\(\Rightarrow P_{min}=\sqrt{17}\) khi \(x=y=\frac{1}{2}\)

\(\frac{y}{x}+\frac{x}{y}\ge2\left(Cauchy\right)\Rightarrow Min=2\Leftrightarrow x=y\)