\(P\left(x\right)=\left(x^2+2.\frac{b}{2}x+\frac{b^2}{4}\right)+c-\frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2+c-\frac{b^2}{4}\ge c-\frac{b^2}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{-b}{2}\)

Mà min P(x)=-1 khi x=2 \(\Rightarrow\)\(\hept{\begin{cases}\frac{-b}{2}=2\\c-\frac{b^2}{4}=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}b=-4\\c=3\end{cases}}\)

\(\Rightarrow\)\(P\left(x\right)=x^2-4x+3\)

Ta có \(\sqrt{3x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+z\right)\left(x+y\right)}\ge\sqrt{xy}+\sqrt{xz}\)(BĐT buniacoxki)

=>\(VT\le\frac{x}{x+\sqrt{xz}+\sqrt{xy}}+\frac{y}{y+\sqrt{yx}+\sqrt{yz}}+\frac{z}{z+\sqrt{zx}+\sqrt{yz}}\)

=> \(VT\le\frac{\sqrt[]{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

#)Giải : 

Ta có : 

\(\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\forall x\)

\(\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=4\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\Leftrightarrow\hept{\begin{cases}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4=2}\\3-\left(x-1\right)^2=3\end{cases}}\)

\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất là x = 1

bạn ơi sao suy ra đc là VT lơn hơn 3