Chủ đề này có 3 trả lời

[dai101001000]

CMR nếu:

$ a, b, c, d, e, f> 0$

$ a+ b+ c+ d+ e+ f= 6$

$ a^{2}+ b^{2}+ c^{2}+ d^{2}+ e^{2}+ f^{2}= \frac{36}{5}$ thì:

$ a^{3}+ b^{3}+ c^{3}+ d^{3}+ e^{3}+ f^{3}\leq  \frac{264}{25}$

[dai101001000]

CMR nếu:

$ a, b, c, d, e, f> 0$

$ a+ b+ c+ d+ e+ f= 6$

$ a^{2}+ b^{2}+ c^{2}+ d^{2}+ e^{2}+ f^{2}= \frac{36}{5}$ thì:

$ a^{3}+ b^{3}+ c^{3}+ d^{3}+ e^{3}+ f^{3}\leq  \frac{264}{25}$

ai giải hộ mình với. Mình cảm ơn!

[DOTOANNANG]

$\left ( 6- a \right )^{2}= \left ( b+ c+ d+ e+ f \right )^{2}\leq 5\left ( b^{2}+ c^{2}+ d^{2}+ e^{2}+ f^{2} \right )= 5\left ( \frac{36}{5}- a^{2} \right )\Leftrightarrow a\leq 2$

$a= 2\Leftrightarrow b= c= d= e= f= \frac{4}{5}$

$\Leftrightarrow a^{3}+ b^{3}+ c^{3}+ d^{3}+ e^{3}+ f^{3}\leq \frac{264}{25}$

[DOTOANNANG]

$5(\frac{36}{5}-a^2)\geq(6-a)^2$, $a \leq 2$, $b,c,d,e,f \leq 2$

$(2-a)(a-\frac{4}{5})^2 \geq 0$, $a^3 \leq \frac{18}{5}a^2-\frac{96}{25}a+\frac{32}{25}$, $\sum a^3 \leq \frac{264}{25}$