# Ôn thi vào 10

Cho $a,b,c>0$ thỏa mãn $ab+bc+ca=3$.Tìm Max:

$P=\dfrac{a}{a^2+4a+3}+\dfrac{b}{b^2+4b+3}+\dfrac{c}{c^2+4c+3}$

9 tháng 4 lúc 5:03

$ab+bc+ca=3\Rightarrow\left\{{}\begin{matrix}a+b+c\ge3\\abc\le1\end{matrix}\right.$

Ta sẽ chứng minh $P\le\dfrac{3}{8}$

$P\le\dfrac{a}{6a+2}+\dfrac{b}{6b+2}+\dfrac{c}{6c+2}$ nên chỉ cần chứng minh: $\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\le\dfrac{3}{4}$

$\Leftrightarrow\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\ge\dfrac{3}{4}$

$\Leftrightarrow\dfrac{\left(3a+1\right)\left(3b+1\right)+\left(3b+1\right)\left(3c+1\right)+\left(3c+1\right)\left(3a+1\right)}{\left(3a+1\right)\left(3b+1\right)\left(3c+1\right)}\ge\dfrac{3}{4}$

$\Leftrightarrow\dfrac{6\left(a+b+c\right)+30}{27abc+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}$

$\Rightarrow\dfrac{6\left(a+b+c\right)+30}{27+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}$

$\Leftrightarrow24\left(a+b+c\right)+120\ge165+9\left(a+b+c\right)$

$\Leftrightarrow a+b+c\ge3$ (đúng)

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