Question

tìm Max của P=\(\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\) biết a+b+c=1

@Lightning Farron

Answer 1

Cauchy-Schwarz dạng ENgel:

\(P=\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\)

\(=\dfrac{1}{2}\cdot3-\left(\dfrac{b}{4a+2b}+\dfrac{c}{4b+2c}+\dfrac{a}{4c+2a}\right)\)

\(=\dfrac{3}{2}-\left(\dfrac{b^2}{4ab+2b^2}+\dfrac{c^2}{4bc+2c^2}+\dfrac{a^2}{4ac+2a^2}\right)\)

\(\le\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}\)

\(=\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=1\)

\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)