Inegalitate a+b+c=3, 5a^2+(b+c)^2+... (own)

[Marius Stănean]

$a,b,c\ge 0,\;a+b+c=3\;\Longrightarrow$

• $\dfrac{5a+2(b+c)}{5a^2+(b+c)^2}+\dfrac{5b+2(c+a)}{5b^2+(c+a)^2}+\dfrac{5c+2(a+b)}{5c^2+(a+b)^2}\le 3$