tìm Max của P=\(\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\) biết a+b+c=1
Cauchy-Schwarz dạng ENgel:
\(P=\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\)
\(=\dfrac{1}{2}\cdot3-\left(\dfrac{b}{4a+2b}+\dfrac{c}{4b+2c}+\dfrac{a}{4c+2a}\right)\)
\(=\dfrac{3}{2}-\left(\dfrac{b^2}{4ab+2b^2}+\dfrac{c^2}{4bc+2c^2}+\dfrac{a^2}{4ac+2a^2}\right)\)
\(\le\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}\)
\(=\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=1\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
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