Vào link này nhé ,mình tìm cả max và min luôn

https://olm.vn/hoi-dap/detail/221940896077.html

Hoặc trong câu hỏi tương tự cũng có 

\(2\sqrt{xy}+\sqrt{2x}+\sqrt{2y}\ge8\)

Mà \(\left\{{}\begin{matrix}2\sqrt{xy}\le x+y\\\sqrt{2x}+\sqrt{2y}\le2\sqrt{x+y}\end{matrix}\right.\)

\(\Rightarrow x+y+2\sqrt{x+y}\ge8\)

\(\Leftrightarrow\left(\sqrt{x+y}-2\right)\left(\sqrt{x+y}+4\right)\ge0\)

\(\Rightarrow x+y\ge4\)

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

\(P\ge\frac{x+y}{4}+\frac{4}{x+y}+\frac{3\left(x+y\right)}{4}\ge2\sqrt{\frac{4\left(x+y\right)}{4\left(x+y\right)}}+\frac{3.4}{4}=5\)

Dấu "=" xảy ra khi \(x=y=2\)

\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}\)

\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}\)

\(VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)\)

\(VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

\(\left\{{}\begin{matrix}x+1=a\\y+1=b\end{matrix}\right.\) \(\Rightarrow a+b=4\)

\(P=\frac{1}{\sqrt{a^2+1}+a}+\frac{1}{\sqrt{b^2+1}+b}=\sqrt{a^2+1}-a+\sqrt{b^2+1}-b\)

\(P=\sqrt{a^2+1}+\sqrt{b^2+1}-4\)

\(P\ge\sqrt{\left(a+b\right)^2+\left(1+1\right)^2}-4=2\sqrt{5}-4\)

\(P_{min}=2\sqrt{5}-4\) khi \(a=b=2\) hay \(x=y=1\)

GTNN

\(x^2+y^2=1=\left(x+y\right)^2-2xy\Rightarrow2xy=\left(x+y\right)^2-1\)

\(x;\text{ }y\ge0\Rightarrow x+y=\sqrt{x^2+y^2+2xy}\ge\sqrt{1+2xy}\ge1\)

\(A^2=2+2\left(x+y\right)+2\sqrt{\left(1+2x\right)\left(1+2y\right)}\)

\(=2+2\left(x+y\right)+2\sqrt{1+2\left(x+y\right)+4xy}\)

\(=2+2\left(x+y\right)+2\sqrt{1+2\left(x+y\right)+2\left(x+y\right)^2-2}\)

\(=2+2t+2\sqrt{2t^2+2t-1}\text{ }\left(t=x+y\ge1\right)\)

\(\ge2+2+2\sqrt{2.1^2+2.1-1}\)

\(=4+2\sqrt{3}\)

\(\Rightarrow A\ge\sqrt{4+2\sqrt{3}}=1+\sqrt{3}\)

Dấu bằng xảy ra khi \(x+y=1\Leftrightarrow xy=0\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

GTLN

Với 2 số thực bất kì, ta luôn có: \(\left(a+b\right)^2=2\left(a^2+b^2\right)-\left(a-b\right)^2\le2\left(a^2+b^2\right)\)

\(A^2\le2\left(1+2x+1+2y\right)=4+4\left(x+y\right)\le4+4\sqrt{2\left(x^2+y^2\right)}=4+4\sqrt{2}\)

\(\Rightarrow A\le\sqrt{4+4\sqrt{2}}\)

Dấu bằng xảy ra khi 2 biến bằng nhau.

5(x+y)²+3(x-y)²=8x²+4xy+8y²=4(2x²+xy+2z²)>=5(x+y)²

^{=> \(\sqrt{2x^2+xy+2y^2}\ge\sqrt{\frac{5\left(x+y\right)^2}{4}}\)= \(\frac{\sqrt{5}\left(x+y\right)}{2}\)}

^{Tương tự. Cộng lại là ra nha. Dấu = xảy ra <=> x=y=z=1/3}