\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)

Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)

tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)

=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)

Dấu "=" xảy ra khi x=y=z=4

Vậy minM=6 khi x=y=z=4

b1: Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

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

=>minP=1 <=> x=y=z=2/3

Dễ dàng CM được BĐT sau: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)(BĐT Nestbit)

Vậy: \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge3\)

\(\Leftrightarrow P+a+b+c\ge3\Leftrightarrow P\ge3-2=1\)

Vậy Min P=1 <=> x=y=z=\(\frac{2}{3}\)

ta có: \(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)(dấu = xảy ra khi \(\left(y+z\right)^2=4x^2\)↔y+z=2x)

tương tự ta có:\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y;\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)(dấu = cũng xảy ra khi x+z=2y;x+y=2z)

cộng từng vế ta có:P+\(\frac{x+y+z}{2}\ge x+y+z\)

→P\(\ge\frac{x+y+z}{2}\)mà x+y+x=1

\(P\ge\frac{1}{2}\)↔\(\begin{cases}y+z=2x\\x+z=2y\\x+y=2z\end{cases}\)→x=y=z=1/3

Áp dụng BĐT cauchy schawrz dạng engel ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Áp dụng BĐT cauchy schawrz dạng engel, ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Áp dụng bất đẳng thức Svacxo ta có : 

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

\(=\frac{\left(2x+2y+2z\right)^2}{x+y+z}=\frac{\left[2\left(x+y+z\right)\right]^2}{x+y+z}=\frac{4\left(x+y+z\right)^2}{x+y+z}=4\left(x+y+z\right)\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z\)

Vậy ta có điều phải chứng minh

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Áp dụng bất đẳng thức svác sơ ta có 

\(A\ge\frac{\left(x+y+z\right)^2}{y+3z+z+3x+x+3y}=\frac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+x}{4}=\frac{3}{4}\)

Đặt \(P=\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\)

Áp dụng bất đẳng thức Canchy Schwarz dạng Engel : 

\(P=\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}>\frac{\left(x+y+z\right)^2}{y+3y+z+3z+x+3x}=\frac{\left(x+y+z\right)^2}{4x+4y+4z}=\frac{\left(x+y+z\right)^2}{4.\left(x+y+z\right)}=\frac{3^2}{4}=\frac{3}{4}\)

Dấu " = " xảy ra khi x=y=z=1.

Sử dụng AM - GM ta dễ có:

\(\frac{x^2}{y+3z}+\frac{y+3z}{16}\ge2\sqrt{\frac{x^2}{y+3z}\cdot\frac{y+3z}{16}}=\frac{x}{2}\)

Tương tự:

\(\frac{y^2}{z+3x}+\frac{z+3x}{16}\ge\frac{y}{2};\frac{z^2}{x+3y}+\frac{x+3y}{16}\ge\frac{z}{2}\)

Khi đó:

\(\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\ge\frac{x+y+z}{2}-\frac{x+y+z}{4}=\frac{x+y+z}{4}=\frac{3}{4}\)

Đẳng thức xảy ra tại x=y=z=1