For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$

From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But clearly, 15 can be less than $5(a+b+c)$. So this technique won't work. I fail to see some other inequalities to prove this. Any other ideas/hints? Thanks

• set $a=x/y,b=y/z,c=z/x$ – Dr. Sonnhard Graubner Dec 25 '15 at 2:12
• No. $5(a+b+c)$ cannot be less than $15$. – SinTan1729 Dec 25 '15 at 2:35
• @SayantanSantra The question has been edited. – user236182 Dec 25 '15 at 7:45
• By AM-GM $5(a+b+c)\ge 5(3\sqrt[3]{abc})=15$. So in fact, $15$ is always less than $5(a+b+c)$ (or equal if $a=b=c=1$). – user236182 Dec 25 '15 at 7:46

Lemma: for any postive $x,y,z>0$,and $xyz=1$ we have $$(xy+yz+xz)^2+3\ge 4xyz(x+y+z)$$ proof:since Use Schur inequality we have $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)\ge 2((ab)^{3/2}+(bc)^{3/2}+(ac)^{3/2})$$ take $a'^2=a^3,b'^2=b^3,c'^2=c^3$ and Assmue that $a'b'c'=1$so we have $$a'^2+b'^2+c'^2+3\ge 2(a'b'+b'c'+c'a'),$$ so $$(a'+b'+c')^2+3\ge 4(a'b'+b'c'+c'a')$$ Let $a'=xy,b'=yz,c'=zx$ by done.

let $a+b+c=p,ab+bc+ac=q,abc=r=1$,your inequality equal to $$pq+6\ge 5p$$ Use Lemma we have $$q^2+3\ge 4p$$ $$\Longleftrightarrow p\sqrt{4p-3}+6\ge 5p,p\ge 3$$ it is clear .Because $$p^2(4p-3)-(5p-6)^2=4(p-3)^2(p-1)\ge 0$$

• Thank you very much, appreciated – amathnerd Dec 25 '15 at 17:06

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality is a linear inequality of $u$, which says that it's enough to prove our inequality

for an extremal value of $u$, which happens for equality case of two variables.

Let $b=a$.

Thus, $z=\frac{1}{a^2}$ and we need to prove that $$(a-1)^2(2a^4+4a^3-4a^2-a+2)\geq0,$$ which is obvious.

Done!