Tìm GTNN và GTLN nâng cao.

E

eye_smile

1,$36=(a+b+c)^2=(a+\dfrac{1}{2}.2b+\dfrac{1}{3}.3c)^2 \le (1+\dfrac{1}{4}+\dfrac{1}{9})(a^2+4b^2+9c^2)= \dfrac{49}{36}.(a^2+4b^2+9c^2)$

\Rightarrow $a^2+4b^2+9c^2 \ge \dfrac{36^2}{49}$

2,$1=(a+b+c)^2=(a+\dfrac{1}{\sqrt{2}}.\sqrt{2}b+ \dfrac{1}{\sqrt{3}}. \sqrt{3}c)^2 \le (1+\dfrac{1}{2}+\dfrac{1}{3})(a^2+2b^2+3c^2)$

\Rightarrow $a^2+2b^2+3c^2 \ge \dfrac{6}{11}$
 
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