Question

Cho 3 số thực dương abc

CMR \(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}=\frac{a^2+c^2}{a+c}\le\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Answer 1

Lời giải:

Sử dụng pp biến đổi tương đương. Ta có:

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

\(\Leftrightarrow \left(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}\right)(a+b+c)\leq 3(a^2+b^2+c^2)\)

\(\Leftrightarrow a^2+b^2+\frac{c(a^2+b^2)}{a+b}+b^2+c^2+\frac{a(b^2+c^2)}{b+c}+c^2+a^2+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)\)

\(\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}-(a^2+b^2+c^2)\leq 0\)

\(\Leftrightarrow \frac{c(a^2+b^2)}{a+b}-c^2+\frac{a(b^2+c^2)}{b+c}-a^2+\frac{b(a^2+c^2)}{a+c}-b^2\leq 0\)

\(\Leftrightarrow \frac{ac(a-c)+bc(b-c)}{a+b}+\frac{ab(b-a)+ac(c-a)}{b+c}+\frac{ab(a-b)+bc(c-b)}{a+c}\leq 0\)

\(\Leftrightarrow ac(c-a)\left(\frac{1}{b+c}-\frac{1}{a+b}\right)+bc(b-c)\left(\frac{1}{a+b}-\frac{1}{a+c}\right)+ab(a-b)\left(\frac{1}{a+c}-\frac{1}{b+c}\right)\leq 0\)

\(\Leftrightarrow \frac{-ac(c-a)^2}{(b+c)(a+b)}+\frac{-bc(b-c)^2}{(a+b)(a+c)}+\frac{-ab(a-b)^2}{(a+c)(b+c)}\leq 0\)

\(\Leftrightarrow \frac{ac(c-a)^2}{(b+c)(a+b)}+\frac{bc(b-c)^2}{(a+b)(a+c)}+\frac{ab(a-b)^2}{(a+c)(b+c)}\geq 0\) (luôn đúng với mọi $a,b,c>0$)

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Answer 2

bai lam don gian 3 dong

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

\(\Leftrightarrow3(a^2+b^2+c^2)(a+b)(b+c)(c+a)\ge(a+b+c)\left(\sum_{cyc}(a^2+b^2)(c+a)(c+b)\right)\)

\(\Leftrightarrow\sum_{perms}a^2b(a-b)^2\ge0\)