Question

Let a,b,c non-negativenumber such that a+b+c=3. prove that:
1/(2ab^2 +1) + 1/(2bc^2 +1) + 1/(2ca^2 +1) >= 1

Answer 1

what u means mukushla??

Answer 2

i mean i'll post my answer later
i gotta go now...

Answer 3

i dont wanna make a complete solution
try to work it out

1.

\[ \frac{1}{2ab^{2}+1}+\frac{1}{2bc^{2}+1}+\frac{1}{2ca^{2}+1}\geq 1\:\Longleftrightarrow\:\frac{ab^{2}}{2ab^{2}+1}+\frac{bc^{2}}{2bc^{2}+1}+\frac{ca^{2}}{2ca^{2}+1}\leq 1 \]
(why?)


2. By AM-GM we have \(2ab^{2}+1\geq 3\sqrt[3]{a^{2}b^{4}}\) hence inequality becomes to
\[ \frac{ab^{2}}{2ab^{2}+1}+\frac{bc^{2}}{2bc^{2}+1}+\frac{ca^{2}}{2ca^{2}+1}\leq\frac{1}{3}(\sqrt[3]{ab^{2}}+\sqrt[3]{bc^{2}}+\sqrt[3]{ca^{2}}) \]
(why?)


3.and Using Chebyshev gives

\[ \frac{1}{3}(\sqrt[3]{ab^{2}}+\sqrt[3]{bc^{2}}+\sqrt[3]{ca^{2}})\leq\frac{a+b+c}{3}= 1 \]
(why?)

Answer 4

U can study about Chebyshev's inequality and some other inequalities from this usefull resource

Answer 5

OK, million thanks @mukushla...

Answer 6

yw :)