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Question

Cho các số thuc duong a,b,c thoa man:

$a^{2}+b^{2}+c^{2}=14$

Tim gia tri lon nhat cua:

$P=\frac{4(a+c)}{a^{2}+3c^{2}+28}+\frac{4a}{a^{2}+bc+7}-\frac{5}{(a+b)^{2}}-\frac{3}{a(b+c)}$

WhjteShadow · Answer 1 · 11/10/2014 6:34:00 PM

Có

$14+2bc=a^2+(b+c)^2\geq 2a(b+c)\Rightarrow a(b+c)\leq7+bc$

$\Rightarrow \frac{4a}{a^2+bc+7}\leq \frac{4}{a+b+c}$

$a(b+c)\leq (\frac{a+b+c}{2})^2$

$\Rightarrow -\frac{3}{a(b+c)}\leq -\frac{12}{(a+b+c)^2}$

$\frac{4}{a+b+c}-\frac{12}{(a+b+c)^2} \leq \frac{1}{3}\Rightarrow \frac{4a}{a^2+bc+7}-\frac{3}{a^2+bc+7}\leq \frac{1}{3}$

$(1+\frac{1}{3})(a^2+3c^2) \geq (a+b)^2\Rightarrow a^2+3c^2\geq \frac{3}{4}(a+b)^2$

$\Rightarrow \frac{4(a+c)}{a^2+3c^2+28}\leq \frac{4(a+c)}{\frac{3}{4}(a+c)^2+28}\leq \frac{1}{25}(a+c-4)+\frac{2}{5}$

$-\frac{5}{(a+b)^2}\leq \frac{2}{25}(a+b-5)-\frac{1}{5}$

$\Rightarrow \frac{4(a+c)}{a^2+3c^2+28}-\frac{5}{(a+b)^2}\leq \frac{3a+2b+c-9}{25}$

$(3a+2b+c)\leq \sqrt{(a^2+b^2+c^2)(3^2+2^2+1^2)}$

Từ đó

$Max=\frac{8}{15}$

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