Dạng 3: Kỹ thuật chọn điểm rơi trong bài toán cực trị đạt được tại tâm có đáp án

$A=\left(4a+4b+\frac{1}{a}+\frac{1}{b}\right)-3a-3b\ge 4\sqrt[4]{4a\mathrm{..4}b.\frac{1}{a}.\frac{1}{b}}-3\left(a+b\right)\ge 8-3=5$

$\begin{array}{c}A=\left(a^2+b^2+c^2+\frac{1}{8a}+\frac{1}{8b}+\frac{1}{8c}+\frac{1}{8a}+\frac{1}{8b}+\frac{1}{8c}\right)+\frac{3}{4a}+\frac{3}{4b}+\frac{3}{4c}\\ \ge 9\sqrt[9]{a^2.b^2.c^2.\frac{1}{8a}.\frac{1}{8b}.\frac{1}{8c}.\frac{1}{8a}.\frac{1}{8b}.\frac{1}{8c}}+\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ \ge \frac{9}{4}+9.\frac{1}{\sqrt[3]{abc}}\ge \frac{9}{4}+\frac{9}{4}.\frac{1}{\frac{a+b+c}{3}}\ge \frac{9}{4}+\frac{9}{4}.2=\frac{27}{4}\end{array}$

$A=\left(\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\right)+\frac{3\left(a+b\right)}{4\sqrt{ab}}\ge \text{2}\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}+\frac{3.2\sqrt{ab}}{4\sqrt{ab}}=1+\frac{3}{2}=\frac{5}{2}$

Dấu “=” xảy ra $\iff a=b$

$\begin{array}{c}A=\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{b+c}{4a}+\frac{c+a}{4b}+\frac{a+b}{4c}\right)+\frac{3}{4}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\\ \ge 6\sqrt[6]{\frac{a}{b+c}.\frac{b}{c+a}.\frac{c}{a+b}.\frac{b+c}{4a}.\frac{c+a}{4b}.\frac{a+b}{4c}}+\frac{3}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\end{array}$

$\ge 3+\frac{3}{4}\mathrm{.6.}\sqrt[6]{\frac{b}{a}.\frac{c}{a}.\frac{c}{b}.\frac{a}{b}.\frac{a}{c}.\frac{b}{c}}=3+\frac{9}{2}=\frac{15}{2}$

$A=\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge 2\sqrt{\frac{1}{\left(a^2+b^2\right)2ab}}\ge 2.\frac{1}{\frac{a^2+b^2+2ab}{2}}=\frac{4}{{\left(a+b\right)}^2}\ge 4$

$\begin{array}{c}A=\frac{1}{1+a^2+b^2}+\frac{1}{6ab}+\frac{1}{3ab}\\ \ge 2\sqrt{\frac{1}{\left(1+a^2+b^2\right)6ab}}+\frac{1}{3ab}\\ \ge 2.\frac{1}{\frac{1+a^2+b^2+6ab}{2}}+\frac{1}{3ab}=\frac{4}{{\left(a+b\right)}^2+1+4ab}+\frac{1}{3ab}\end{array}$

$\ge \frac{4}{{\left(a+b\right)}^2+1+4{\left(\frac{a+b}{2}\right)}^2}+\frac{1}{3{\left(\frac{a+b}{2}\right)}^2}\left(\text{Do }ab\le {\left(\frac{a+b}{2}\right)}^2\right)$

$\ge \frac{4}{2{\left(a+b\right)}^2+1}+\frac{4}{3{\left(a+b\right)}^2}$

$\ge \frac{4}{2.1+1}+\frac{4}{3.1}=\frac{8}{3}$

$\begin{array}{c}A=\frac{1}{a^2+b^2}+\frac{1}{2ab}+4ab+\frac{1}{4ab}+\frac{1}{4ab}\\ \ge 2\sqrt{\frac{1}{\left(a^2+b^2\right)2ab}}+2\sqrt{4ab.\frac{1}{4ab}}+\frac{1}{4ab}\\ \ge 2.\frac{1}{\frac{a^2+b^2+2ab}{2}}+2+\frac{1}{4ab}=\frac{4}{{\left(a+b\right)}^2}+2+\frac{1}{4ab}\end{array}$

$\ge \frac{4}{{\left(a+b\right)}^2}+2+\frac{1}{4{\left(\frac{a+b}{2}\right)}^2}\left(\text{Do }ab\le {\left(\frac{a+b}{2}\right)}^2\right)$

$\begin{array}{c}\ge \frac{5}{{\left(a+b\right)}^2}+2\\ \ge \frac{5}{1}+2=7\end{array}$