Question

Cho a,b,c > 0. CMR:

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

Answer 1

Hân đz đã đến :v giờ lm nha

Ta có: \(a^3=a\cdot a^2\)

\(\Rightarrow a^3+a\cdot b^2=a\cdot a^2+a\cdot b^2=a\left(a^2+b^2\right)\)

\(\Rightarrow\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\)(*)

Ta có: \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)

\(\Rightarrow\dfrac{ab^2}{a^2+b^2}\le\dfrac{ab^2}{2ab}=\dfrac{b}{2}\)

\(\Rightarrow\dfrac{a^3}{a^2+b^2}\ge a-\dfrac{b}{2}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\); \(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)

Cộng 3 bđt trên ta có:

\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\)

''='' xảy ra khi \(a=b=c\)

Answer 2

lát hông ai làm thì t lm cho :))