jasperE3 wrote: If $a,b,c$ are positive numbers, prove the inequality $$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+c)(b+a)}+\frac{c^2}{(c+a)(c+b)}\ge\frac34.$$ i will make an actual proof with this using AM-GM and other stuff later.

This could be a proof if I could find $\min(a,b,c)$ and subsitute it back into the inequality to show that it works, but im having trouble finding it based off the info given...

jasperE3 wrote: If $a,b,c$ are positive numbers, prove the inequality $$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+c)(b+a)}+\frac{c^2}{(c+a)(c+b)}\ge\frac34.$$ Using Titu's lemma, we have $$\sum_{ \text{cyc} } \frac{a^2}{(a+b)(a+c)} \quad \enspace \geq \quad \enspace \frac{ (\sum_{\text{cyc}} a )^2}{ \left(\sum_{ \text{cyc} }a^2 + 3 \sum_{\text{cyc}} ab \right) } $$ Therefore, it suffices to show that $$ \frac{ (\sum_{\text{cyc}} a )^2}{ \left(\sum_{ \text{cyc} }a^2 + 3 \sum_{\text{cyc}} ab \right) } \quad \geq \quad \frac{3}{4}$$ but this is equivalent to $$ \frac{1}{2} \sum_{ \text{cyc} } (a-b)^2 \quad \geq \quad 0 $$ which is obviously true.

If $a,b,c$ are positive numbers, prove the inequality $$\frac{3(a^2+b^2+c^2)}{4(ab+bc+ca)}\geq \frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+c)(b+a)}+\frac{c^2}{(c+a)(c+b)}\geq\frac34$$$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+c)(b+a)}+\frac{\lambda c^2}{(c+a)(c+b)} \geq1-\frac{1}{4\lambda}.$$Where $\lambda>\frac{1}{2}.$

apply titu's lemma on LHS and then, using cauchy shwarz inequality on (a^2, b^2, c^2) and (b^2, c^2, a^2) and some minor manipulations we get that, -((ab+bc+ca)/(a^2 + b^2 + c^2 + 3(ab+bc+ca)) >= -1/4 now add 1 on both sides of the above inequality