Symmetric inequality over 3 variables with restriction $a+b+c=1$

Question

Suppose that $a,b,c$ are positive real numbers with $a+b+c=1$. Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{2(a^3+b^3+c^3)}{abc}.$$

Since I am texting from cellphone I will skip some details. I multiplied both sides by $abc$ and I got $ab+bc+ac\leq 3abc+2(a^3+b^3+c^3)$. And I expressed everything through elementary symmetric functions $\sigma_i$ taking into account that $\sigma_1=1$. Then it is equivalent that I need to show that $9\sigma_3-7\sigma_2+2\geq 0$. But I have some troublez to show the last inequality.

Can anyone explain the proof, please?

Answer 1

The homogenization helps.

We need to prove that: $$2(a^3+b^3+c^3)+3abc\geq(a+b+c)(ab+ac+bc)$$ or $$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or $$\sum_{cyc}(a^3-a^2b-ab^2+b^3)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)\geq0,$$ which is obvious.

Answer 2

After homogenization, we want to prove $$2(a^3+b^3+c^3)+3abc\geqslant(a+b+c)(ab+ac+bc)$$ $$\iff 2(a^3+b^3+c^3)\geqslant ab(a+b)+bc(b+c)+ca(c+a)$$ Use now the well-known factorization $a^3+b^3=(a+b)(a^2-ab+b^2)\geqslant (a+b)ab$ to end it.

Notice that the last inequality also follows from a simple application of AM-GM: $$\frac{a^3+a^3+b^3}3\geqslant a^2b\implies a^3+b^3\geqslant a^2b+ab^2=ab(a+b)$$ It is also Muirhead for $(3,0)\succ(2,1)$.