Inequality with $a+b+c=1$, a,b,c positive numbers

Question

If a,b,c are real positive numbers, such as $ a+b+c=1$ prove that $ \left( ab+bc+ca\right) \left( 1+3abc\right) \geq 10abc $.

Thank you!

First try:$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(1+3xyz)\geq10$$ witch doesn’t works because I have obtained $xy+yz+zx\geq\frac{1}{3}$ and this is false.

Second try: $$xy+yz+zx\geq9xyz$$ that does not work either

Answer 1

After homogenization we need to prove that $$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2-3a^3bc-5a^2b^2c)\geq0,$$ which is true by Muirhead.

Also, we can use SOS:

We need to prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3(ab+ac+bc)\geq10$$ or $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{9}{a+b+c}\geq(a+b+c)^2-3(ab+ac+bc)$$ or $$\frac{\sum\limits_{cyc}c(a-b)^2}{abc}\geq\sum_{cyc}(a^2-ab)$$ or $$\sum_{cyc}(a-b)^2\left(\frac{1}{ab}-\frac{1}{2}\right)\geq0,$$ which is true because $$2=2(a+b+c)^2>2(a+b)^2>ab.$$