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[LoveMath1234]

Cho $a,b,c>0: a+b+c=ab+bc+ca$

CMR: $\sum \frac{a+b+1}{a^2+b^2+1}\leq 3$

[yeutoan2001]

Có $(a^{2}+b^{2}+1)(1+1+1)\geq (a+b+1)^{2}$

Tương tự =>

   $\sum \frac{a+b+1}{a^{2}+b^{2}+1}\leq 3\sum \frac{1}{a+b+1}\leq 3\sum \frac{(a+b+c^2)}{(a+b+1)(a+b+c^2)}$

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