# If $a+b+c=1$ and a,b,c ＞0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$ [duplicate]

If $$a+b+c=1$$ and a,b,c＞0 prove $$\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?

## marked as duplicate by Martin R, Macavity inequality StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 22 at 8:52

For positive variables by C-S $$\sum_{cyc}\frac{a^2}{a^2+c}=\sum_{cyc}\frac{a^2(a+b)^2}{(c(a+b+c)+a^2)(a+b)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2}.$$ Thus, it's enough to prove that $$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+c^2+ac+bc)(a+b)^2$$ or $$\sum_{cyc}(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)\geq0,$$ which is true because $$\sum_{cyc}(a^4-a^3b)\geq0$$ by Rearrangement; $$\sum_{cyc}a^3c\geq\sum_{cyc}a^2bc$$ it's $$\sum_{cyc}\frac{a^2}{b}\geq\sum_{cyc}a,$$ which is true by Rearrangement again and $$\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$