$1) a,b,c\geq 0; c\leq a\leq b$
Tìm GTNN $S=\frac{1}{a^{2}+c^{2}}+\frac{1}{b^{2}+c^{2}}+\sqrt{a+b+c}$
$2) a,b,c >0; a+b+c=1$
Tìm GTLN $T = \frac{4}{a+b}+\frac{4}{c+b}+\frac{4}{a+c}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$
$2) a,b,c >0; a+b+c=1$
Tìm GTLN $T = \frac{4}{a+b}+\frac{4}{c+b}+\frac{4}{a+c}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$
$2)$
Ta có:
$$\sum \frac{4}{a+b}=\sum \frac{4(a+b+c)}{a+b}=\sum \frac{4c}{a+b}+12\leq \sum (\frac{c}{a}+\frac{c}{b})+12=\sum \frac{a+b}{c}+12(1)$$
Và:
$$\sum \frac{1}{a}=\sum \frac{a+b+c}{a}=\sum \frac{a+b}{c}+3(2)$$
Từ $(1),(2)$ suy ra:
$$T=\sum \frac{4}{a+b}-\sum \frac{1}{a}\leq (\sum \frac{a+b}{c}+12)-(\sum \frac{a+b}{c}+3)=9$$
Dấu đẳng thức xảy ra $\Leftrightarrow a=b=c=\frac{1}{3}$
$\color{red}{\mathrm{\text{How I wish I could recollect, of circle roud}}}$
$\color{red}{\mathrm{\text{The exact relation Archimede unwound ! }}}$
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