giải bpt!

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ovan

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vansang02121998

$\sqrt[3]{a+7b}+\sqrt[3]{b+7c}+\sqrt[3]{c+7a} \le a^3+b^3+c^3+3$

Có $9(x^3+y^3+z^3) \ge (x+y+z)^3$, áp dụng

$9(a^3+b^3+c^3) \ge (a+b+c)^3=27 \leftrightarrow a^3+b^3+c^3+3 \ge 6$

$(\sqrt[3]{a+7b}+\sqrt[3]{b+7c}+\sqrt[3]{c+7a})^3 \le 9(a+7b+b+7c+c+7a) = 216 \leftrightarrow \sqrt[3]{a+7b}+\sqrt[3]{b+7c}+\sqrt[3]{c+7a} \le 6$
 
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vansang02121998

$P=\dfrac{3(x+y+z)}{\sqrt{6(x^2+5)}+\sqrt{6(y^2+5)}+\sqrt{6(z^2+5)}}$

$\leftrightarrow P=\dfrac{\sqrt{6}}{2}.$$\dfrac{x+y+z}{\sqrt{x^2+5}+\sqrt{y^2+5}+\sqrt{z^2+5}}$

có $xy+xz+yz=5$

$\rightarrow P=\dfrac{\sqrt{6}}{2}.\dfrac{x+y+z}{\sqrt{(x+y)(x+z)}+\sqrt{(x+y)(y+z)}+\sqrt{(x+z)(y+z)}}$

có $\sqrt{ab} \le \dfrac{a+b}{2}$

$\rightarrow P \ge \dfrac{\sqrt{6}}{2}.\dfrac{x+y+z}{\dfrac{2x+y+z}{2}+\dfrac{x+2y+z}{2}+\dfrac{x+y+2z}{2}}$

$\leftrightarrow P \ge \dfrac{\sqrt{6}}{2}.\dfrac{2(x+y+z)}{4(x+y+z)}$

$\rightarrow P \ge \dfrac{\sqrt{6}}{4}$

Dấu $"="$ xảy ra khi $x=y=z=\dfrac{\sqrt{15}}{3}$
 
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