\(\)Ta có: \(x+y+z+\sqrt{xyz}=4\)

=>\(y + z = 4 - x - \sqrt{xyz}\)

Ta có: \((4-y)(4-z)=16-4(y+z)+yz\)

=>\((4-y)(4-z) = 16 - 4(4 - x - \sqrt{xyz}) + yz\)

\(= 16 - 16 + 4x + 4\sqrt{xyz} + yz\)

\(= 4x + 4\sqrt{xyz} + yz\)

=>\(x(4-y)(4-z) = x(4x + 4\sqrt{xyz} + yz)\)

\(=4x^2+4x\sqrt{xyz}+xyz=\left(2x+\sqrt{xyz}\right)^2\)

=>\(\sqrt{x(4-y)(4-z)} = \sqrt{(2x + \sqrt{xyz})^2} = 2x + \sqrt{xyz}\)

Chứng minh tương tự, ta sẽ có:

\(\sqrt{y(4-z)(4-x)}=2y+\sqrt{xyz};\sqrt{z(4-x)(4-y)}=2z+\sqrt{xyz}\)

\(\sqrt{x(4-y)(4-z)}+\sqrt{y(4-z)(4-x)}+\sqrt{z(4-x)(4-y)}-\sqrt{xyz}\)

\(=(2x+\sqrt{xyz})+(2y+\sqrt{xyz})+(2z+\sqrt{xyz})-\sqrt{xyz}\)

\(=2x+2y+2z+2\sqrt{xyz}\)

\(=2(x+y+z+\sqrt{xyz})=2\cdot4=8\) (ĐPCM)

Ta có \(2\sqrt{x}\le x+1\)

\(4\sqrt{y-1}\le4+y-1=y+3\)

\(6\sqrt{z-2}\le9+z-2=z+7\)

Cộng vế theo vế ta được

\(2\sqrt{x}+4\sqrt{y-1}+6\sqrt{z-2}\le x+y+z+11\)

Dấu = xảy ra khi x = 1, y = 5, z = 11

Ta có: \(\sqrt{a^2-ab+b^2}=\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{1}{4}\left(a+b\right)^2}=\frac{1}{2}\left(a+b\right)\)

khi đó:

\(P\le\frac{1}{\frac{1}{2}\left(a+b\right)}+\frac{1}{\frac{1}{2}\left(b+c\right)}+\frac{1}{\frac{1}{2}\left(a+c\right)}\)

\(=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)

Lại có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{a+b}\)=> \(\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

=> \(P\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{2}\left(\frac{1}{c}+\frac{1}{a}\right)\)

\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

Dấu "=" xảy ra <=> a = b = c = 1

Vậy max P = 3 tại a = b = c =1.

Không thích làm cách này đâu nhưng đường cùng rồi nên thua-_-

Đặt \(\sqrt{x+y}=a;\sqrt{y+z}=b;\sqrt{z+x}=c\) suy ra

\(x=\frac{a^2+c^2-b^2}{2};y=\frac{a^2+b^2-c^2}{2};z=\frac{b^2+c^2-a^2}{2}\). Ta cần chứng minh:

\(abc\left(a+b+c\right)\ge\left(a+b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

Đây là bất đẳng thức Schur bậc 3, ta có đpcm.