Question

[Huỳnh Thiện]

Cho  \(a>b>c>0\) và \(a^2+b^2+c^2=1\)

\(CMR:\)\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\ge\frac{1}{2}\)

Answer 1

[Trần Hippo]

Do a,b,c đối xứng , giả sử \(a\ge b\ge c\) \(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)

Áp dụng BĐT Trư - bê - sép , ta có :

\(a^2.\frac{a}{b+c}+b^2.\frac{b}{a+c}+c^2.\frac{c}{b+c}\ge\frac{a^3+b^3+c^3}{3}.\left(\frac{a}{b+C}+\frac{b}{a+c}+\frac{c}{a+b}\right)=\frac{1}{3}.\frac{3}{2}=\frac{1}{2}\)

\(vậy\) \(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\ge\frac{1}{2}\)( Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Answer 2

[Riio Riyuko]

Chebyshev như vầy nhé : 

Ta có : 

\(3.\Sigma\left(a^2.\frac{a}{b+c}\right)\ge\left(a^2+b^2+c^2\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+c}\right)=\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

Áp dụng bất đẳng thức Nesbit , ta có :

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

Suy ra : \(3.\Sigma\left(a^2.\frac{a}{b+c}\right)\ge\frac{3}{2}\)

<=> \(\Sigma\left(a^2.\frac{a}{b+c}\right)\ge\frac{1}{2}\)

Đẳng thức xảy ra <=> a = b = c = \(\frac{1}{\sqrt{3}}\)