<s>[INST] Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality: 
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers. 
Thanks :) 
 [/INST] By the Cauchy-Schwarz Inequality: $$\frac{a^4}{a^2+ab} + \frac{b^4}{b^2 + bc} + \frac{c^4}{c^2 + ac} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc}$$
Now, I claim that $$\frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc} \ge \frac{ab+ac+bc}{2}$$
Expanding, it suffices to show: $$2a^4 + 2b^4 + 2c^4 + 4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b + 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$
Using the inequality $\frac{3}{4}a^4 + \frac{1}{4}b^4 \ge a^3b$ which is a consequence of AM-GM, we can show: $$2a^4 + 2b^4 + 2c^4 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$
Thus it suffices to show: $$4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$
Which is simply just: $$3a^2b^2 + 3a^2c^2 + 3b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab$$
By AM-GM one has $\frac{1}{2}a^2b^2 + \frac{1}{2}a^2c^2 \ge a^2bc$. From applying this a bunch one easily gets the desired result.
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