P = a$\sqrt[]{a+2b}$ + b$\sqrt[]{b+2c}$ + c$\sqrt[]{c+2a}$
Áp dụng BĐT Cô-si:
a$\sqrt[]{(a+2b)*1}$ ≤ a*$\frac{a+2b+1}{2}$ = $\frac{a^2+2ab+a}{2}$
Chứng minh tương tự:
b$\sqrt[]{(b+2c)*1}$ ≤ $\frac{b^2+2bc+b}{2}$
c$\sqrt[]{(c+2a)*1}$ ≤ $\frac{c^2+2ca+c}{2}$
⇒ A ≤ $\frac{a^2+2ab+a}{2}$ + $\frac{b^2+2bc+b}{2}$ + $\frac{c^2+2ca+c}{2}$
A≤ $\frac{a^2+b^2+c^2+2ab+2bc+2ca+a+b+c}{2}$
A≤$\frac{(a+b+c)^2+(a+b+c)}{2}$
A≤ 1 ( thay a+b+c=1)
Dấu "=" xảy ra ⇔ a+2b=b+2c=c+2a=1=a+b+c
⇔ a=b=c
⇔ a=b=c= 1/3
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