Question

Cho a, b, c > 0. 

Chứng minh : \(\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}\ge\frac{a^3+b^3+c^3}{2}\)

 

 

Answer 1

Dat \(P=\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}\)

\(=\frac{a^6}{a^2b+ca^2}+\frac{b^6}{b^2c+ab^2}+\frac{c^6}{c^2a+bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)

Ta di chung minh:

\(\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\ge\frac{a^3+b^3+c^3}{2}\)

\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)

Ta co BDT:

\(a^3+b^3\ge ab\left(a+b\right)\)

\(b^3+c^3\ge bc\left(b+c\right)\)

\(c^3+a^3\ge ca\left(c+a\right)\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)

Suy ra BDT da duoc chung minh

Dau '=' ra khi \(a=b=c\)

Answer 2

Dat P=\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}P=b+ca4​+c+ab4​+a+bc4​

=\frac{a^6}{a^2b+ca^2}+\frac{b^6}{b^2c+ab^2}+\frac{c^6}{c^2a+bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}=a2b+ca2a6​+b2c+ab2b6​+c2a+bc2c6​≥ab(a+b)+bc(b+c)+ca(c+a)(a3+b3+c3)2​

Ta di chung minh:

\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\ge\frac{a^3+b^3+c^3}{2}ab(a+b)+bc(b+c)+ca(c+a)(a3+b3+c3)2​≥2a3+b3+c3​

\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)⇔2(a3+b3+c3)≥ab(a+b)+bc(b+c)+ca(c+a)

Ta co BDT:

a^3+b^3\ge ab\left(a+b\right)a3+b3≥ab(a+b)

b^3+c^3\ge bc\left(b+c\right)b3+c3≥bc(b+c)

c^3+a^3\ge ca\left(c+a\right)c3+a3≥ca(c+a)

\Rightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)⇒2(a3+b3+c3)≥ab(a+b)+bc(b+c)+ca(c+a)

Suy ra BDT da duoc chung minh

Dau '=' ra khi a=b=ca=b=c