Dạng 1: Bất đẳng thức có đáp án

Áp dụng bất đẳng thức AM - GM, ta có:

$\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc$ (đpcm)

Áp dụng bất đẳng thức AM - GM, ta có:

 $\begin{array}{c}\frac{\sqrt{ac}+\sqrt{bd}}{\sqrt{\left(a+b\right)\left(c+d\right)}}=\sqrt{\frac{a}{\left(a+b\right)}.\frac{c}{\left(c+d\right)}}+\sqrt{\frac{b}{\left(a+b\right)}.\frac{d}{\left(c+d\right)}}\\ \le \frac{1}{2}\left(\frac{a}{a+b}+\frac{c}{c+d}\right)+\frac{1}{2}\left(\frac{b}{a+b}+\frac{d}{c+d}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{c+d}{c+d}\right)=1\end{array}$

$\Rightarrow \sqrt{ac}+\sqrt{bd}\le \sqrt{\left(a+b\right)\left(c+d\right)}$ (đpcm)

Áp dụng bất đẳng thức AM - GM, ta có: $\begin{array}{c}\frac{\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}}{\sqrt{ab}}=\sqrt{\frac{c}{b}.\frac{\left(a-c\right)}{a}}+\sqrt{\frac{c}{a}.\frac{\left(b-c\right)}{b}}\\ \le \frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}\right)+\frac{1}{2}\left(\frac{c}{a}+\frac{b-c}{b}\right)\\ \le \frac{1}{2}\left(\frac{c}{b}+1-\frac{c}{a}\right)+\frac{1}{2}\left(\frac{c}{a}+1-\frac{c}{b}\right)=1\end{array}$

 $\Rightarrow \sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le \sqrt{ab}$ (đpcm)

Áp dụng bất đẳng thức AM - GM, ta có: $a\sqrt{b-1}=\sqrt{a}\sqrt{ab-a}\le \frac{1}{2}\left(a+ab-a\right)=\frac{ab}{2}$ (1)

Tương tự:  $b\sqrt{a-1}\le \frac{ab}{2}$  (2)

Cộng theo vế (1) và (2), ta được:  $a\sqrt{b-1}+b\sqrt{a-1}\le ab$ (đpcm)

Áp dụng bất đẳng thức AM - GM, ta có:

$16ab{\left(a-b\right)}^2=4.\left(4ab\right){\left(a-b\right)}^2\le 4.{\left[\frac{4ab+{\left(a-b\right)}^2}{2}\right]}^2=4.{\left[\frac{{\left(a+b\right)}^2}{2}\right]}^2={\left(a+b\right)}^4$ (đpcm)

Áp dụng bất đẳng thức AM - GM, ta có:

$ab+\frac{a}{b}+\frac{b}{a}=\left(\frac{ab}{2}+\frac{a}{2b}\right)+\left(\frac{ab}{2}+\frac{b}{2a}\right)+\left(\frac{a}{2b}+\frac{b}{2a}\right)\ge 2\sqrt{\frac{ab}{2}.\frac{a}{2b}}+2\sqrt{\frac{ab}{2}.\frac{b}{2a}}+2\sqrt{\frac{a}{2b}.\frac{b}{2a}}=a+b+1$(đpcm)

 Vì $a,b>0$  nên   $\frac{a}{b}>0,\frac{b}{a}>0$

Áp dụng bất đẳng thức AM - GM ta có:

$\frac{a}{b}+\frac{b}{a}\ge 2\sqrt{\frac{a}{b}.\frac{b}{a}}=2$  (đpcm)

Áp dụng bất đẳng thức AM - GM ta có:

$\frac{a^2+2}{\sqrt{a^2+1}}=\frac{a^2+1+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\frac{1}{\sqrt{a^2+1}}\ge 2\sqrt{\sqrt{a^2+1}\frac{1}{\sqrt{a^2+1}}}=2$  (đpcm)

Với  $\forall a\ne 0$, áp dụng bất đẳng thức AM - GM ta có: $\frac{3a^2}{1+9a^4}=\frac{1}{\frac{1}{3a^2}+\frac{9a^4}{3a^2}}=\frac{1}{\frac{1}{3a^2}+3a^2}\le \frac{1}{2\sqrt{\frac{1}{3a^2}.3a^2}}=\frac{1}{2}$

 (đpcm)

 $\begin{array}{c}A={\left(a+1\right)}^2+{\left(\frac{a^2+2a+2}{a+1}\right)}^2\\ ={\left(a+1\right)}^2+{\left[\frac{{\left(a+1\right)}^2+1}{a+1}\right]}^2\\ ={\left(a+1\right)}^2+{\left(a+1+\frac{1}{a+1}\right)}^2\\ =2{\left(a+1\right)}^2+\frac{1}{{\left(a+1\right)}^2}+2\stackrel{Cauchy}{}2\sqrt{2{\left(a+1\right)}^2\frac{1}{{\left(a+1\right)}^2}}+2=2\sqrt{2}+2\end{array}$

Dấu “=” xảy ra khi và chỉ khi $2{\left(a+1\right)}^2=\frac{1}{{\left(a+1\right)}^2}$ hay $a=\frac{-2\pm \sqrt[4]{8}}{2}$

Vậy GTNN của $A=2\sqrt{2}+2$

Áp dụng bất đẳng thức AM - GM ta có: 

$a+\frac{1}{b\left(a-b\right)}=b+\left(a-b\right)+\frac{1}{b\left(a-b\right)}\ge 3\sqrt[3]{b.\left(a-b\right).\frac{1}{b\left(a-b\right)}}=3$

Ta có:

Ta có:

$\begin{array}{l}\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=\frac{1}{2}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}\right)+\frac{1}{2}\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+\frac{1}{2}\left(\frac{c^2}{a^2}+\frac{a^2}{b^2}\right)\\ \ge \sqrt{\frac{a^2}{b^2}\frac{b^2}{c^2}}+\sqrt{\frac{b^2}{c^2}.\frac{c^2}{a^2}}+\sqrt{\frac{c^2}{a^2}.\frac{a^2}{b^2}}=\left|\frac{b}{a}\right|+\left|\frac{c}{b}\right|+\left|\frac{a}{c}\right|\ge \frac{b}{a}+\frac{c}{b}+\frac{a}{c}\end{array}$

Ta có:

$\begin{array}{c}\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\left(1+\frac{b+c}{a}\right)+\left(1+\frac{c+a}{b}\right)+\left(1+\frac{a+b}{c}\right)-3\\ =\frac{a+b+c}{a}+\frac{b+c+a}{b}+\frac{c+a+b}{c}-3\\ =\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3\ge 9-3=6\end{array}$