Đặt vế trái là P và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=4\)

Ta cần chứng minh: \(P=\frac{1}{xy+2yz+zx}+\frac{1}{xy+yz+2zx}+\frac{1}{2xy+yz+zx}\le\frac{1}{xyz}\)

\(P=\frac{1}{xy+yz+yz+zx}+\frac{1}{xy+yz+zx+zx}+\frac{1}{xy+xy+yz+zx}\)

\(P\le\frac{1}{16}\left(\frac{1}{xy}+\frac{2}{yz}+\frac{1}{zx}+\frac{1}{xy}+\frac{1}{yz}+\frac{2}{zx}+\frac{2}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{4}\left(\frac{x+y+z}{xyz}\right)=\frac{1}{4}.\frac{4}{xyz}=\frac{1}{xyz}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=\frac{4}{3}\) hay \(a=b=c=\frac{16}{9}\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\end{matrix}\right.\)\(\Rightarrow x+y+z=xyz\)

\(\Rightarrow P=xy+yz+xz-\sqrt{x^2+1}-\sqrt{y^2+1}-\sqrt{z^2+1}\)

Khi \(a=b=c=\frac{1}{\sqrt{3}}\Rightarrow x=y=z=\sqrt{3}\Rightarrow P=3\)

Ta sẽ chứng minh \(P=3\) là giá tri nhỏ nhất của \(P\)

\(\Rightarrow BDT\Leftrightarrow xy+yz+xz-3\ge\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\)

Ta có BĐT \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2\)

\(\Leftrightarrow\left(xy+yz+xz\right)^2\ge x^2y^2z^2+2xyz\left(x+y+z\right)\)\(=3\left(x+y+z\right)^2\)

Xét \(VT^2=\left(xy+yz+xz-3\right)^2=\left(xy+yz+xz\right)^2-6\left(xy+yz+xz\right)+9\)

\(\ge3\left(x+y+z\right)^2-6\left(xy+yz+xz\right)+9\)\(=3\left(x^2+y^2+z^2\right)+9\left(1\right)\)

Và \(VP^2\le\left(1+1+1\right)\left(x^2+y^2+z^2+3\right)=3\left(x^2+y^2+z^2\right)+9\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM. Vậy \(P_{min}=3\Rightarrow a=b=c=\frac{1}{\sqrt{3}}\)

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

BĐT cần chứng minh tương đương với:

\(c\ge2\sqrt{ab}c+c^2\)

\(\Leftrightarrow c+2\sqrt{ab}\le1\) (Do c > 0).

BĐT trên luôn đúng vì: \(c+2\sqrt{ab}\le_{Cauchy}c+a+b=1\).

Vậy ta có đpcm.

\(c+ab=c\left(a+b+c\right)+ab=\left(c+a\right)\left(c+b\right)\)

\(=\left(\sqrt{c}^2+\sqrt{a}^2\right)\left(\sqrt{c}^2+\sqrt{b}^2\right)\ge\left(c+\sqrt{ab}\right)^2\)

Lời giải:

Do $a+b+c=1$ nên:

\(\text{VT}=\sqrt{\frac{ab}{c(a+b+c)+ab}}+\sqrt{\frac{bc}{a(a+b+c)+bc}}+\sqrt{\frac{ca}{b(a+b+c)+ac}}\)

\(=\sqrt{\frac{ab}{(c+a)(c+b)}}+\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ca}{(b+c)(b+a)}}\)

Áp dụng BĐT AM-GM:

\(\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

\(\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)

\(\sqrt{\frac{ca}{(b+c)(b+a)}}\leq \frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)

Cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

để ý cái này: \(\sum\dfrac{a}{a+2\sqrt{bc}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}}=1\)