¸.·’*★Unnamed★secret.·’*★*¸.·’
For all nonnegative real numbers $a,b$ and $c,$ no two of which aer zero$.$Prove that: $\
color{blue}{\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\geq \frac{3\sqrt{3abc(a+b+c)}(a+b+c)^2}{4(ab+bc+ca)^3
}}$
Bất đẳng thức
¸.·’*★Unnamed★secret.·’*★*¸.·’
For all nonnegative real numbers $a,b$ and $c,$ no two of which aer zero$.$Prove that: $\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\geq \frac{3\sqrt{3abc(a+b+c)}(a+b+c)^2}{4(ab+bc+ca)^3}$
Bất đẳng thức
¸.·’*★Unnamed★secret.·’*★*¸.·’
For all nonnegative real numbers $a,b$ and $c,$ no two of which aer zero$.$Prove that: $\
color{blue}{\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\geq \frac{3\sqrt{3abc(a+b+c)}(a+b+c)^2}{4(ab+bc+ca)^3
}}$
Bất đẳng thức