chứng minh bất đẳng thức

B

braga

[TEX]\large \blue (a+b+c)^2=\(\sqrt[4]{a^3}.\sqrt[4]{a}+\sqrt[4]{b^3}.\sqrt[4]{b}+\sqrt[4]{c^3}.\sqrt[4]{c}\)^2 \\ \le \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}\)\(\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}\) \\ \le \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}\)\(\frac{a+1+1+1}{4}+\frac{b+1+1+1}{4}+\frac{c+1+1+1}{4}\) \\ \le \frac{a+b+c+9}{4} \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}\) \\ \Rightarrow \sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}\geq \frac{(a+b+c)^2}{\frac{a+b+c+9}{4}}=\frac{64}{13}>2\sqrt{2}[/TEX]
 
E

eye_smile

Cách khác:

Ta có :$\sqrt[4]{a^3}>\dfrac{a}{\sqrt{2}}$

\Leftrightarrow $a^3>\dfrac{a^4}{4}$

\Leftrightarrow $4>a$ (luôn đúng)

Tương tự, có: $\sqrt[4]{b^3}>\dfrac{b}{\sqrt{2}}$

$\sqrt[4]{c^3}>\dfrac{c}{\sqrt{2}}$

Cộng theo vế, đc:

$\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}>\dfrac{a+b+c}{\sqrt{2}}=2\sqrt{2}$
 
H

huynhbachkhoa23

Theo Holder:

$\sum \sqrt[4]{a^3} \ge \sqrt[4]{3}\sqrt[4]{(\sum a)^3}=\sqrt[4]{3}\sqrt[4]{64}>\sqrt[4]{64}=2\sqrt{2}$
 
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