\(ĐKXĐ:x\ge0\)

\(\sqrt{x}+6\)luôn lớn hơn 4 thỏa mãn ĐKXĐ:

Vậy \(x\ge0\) là giá trị cần tìm

\(\sqrt{x}+6>4\Rightarrow\sqrt{x}>-2\)

Luôn đúng với mọi giá trị của x 

Ta có \(VT=\frac{a^2}{a\sqrt{b}}+\frac{b^2}{b\sqrt{c}}+\frac{c^2}{c\sqrt{a}}\ge\frac{\left(a+b+c\right)^2}{a\sqrt{b}+b\sqrt{c}+c\sqrt{a}}\)

Mà \(a\sqrt{b}\le\frac{a^2+b}{2},b\sqrt{c}\le\frac{b^2+c}{2},c\sqrt{a}\le\frac{c^2+a}{2}\)

=> \(VT\ge\frac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+a+b+c}=\frac{2\left(a+b+c\right)^2}{3+a+b+c}\)

Lại có \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

=> \(VT\ge\frac{2\left(a+b+c\right)^2}{3+3}=\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(=\frac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=2\sqrt{b}\)

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{-b+\sqrt{a}.\sqrt{b}}{\sqrt{b}}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right].\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right).\sqrt{b}}-\frac{\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right]-\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\sqrt{a}+2b.\sqrt{b}}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=2\sqrt{b}\)