Proving $\frac ab+\frac bc+\frac ca\geqslant a+b+c$ when $a$, $b$, $c$ are positive real numbers and $abc=1$

Question

If $a$, $b$, $c$ are positive real number and abc=1. Then prove that $$\frac ab+\frac bc+\frac ca\geqslant a+b+c$$

Using AM-GM both sides will be greater than $3$, But I don't know how to proceed further. Which inequality should I use?

Answer 1

An answer with my thought process.

We want to use $abc=1$, so we clear the denominator on the left:

$a^2c+ab^2+c^2b \ge a+b+c$

Now we are scared because we have something cyclic on the left, something symmetric on the right, and still inhomogenous. We decide to homogenize, so that we can use inequalities like AM-GM:

$a^2c+ab^2+c^2b \ge (a+b+c)(abc)^{2/3}=a^{5/3}b^{2/3}c^{2/3}+a^{2/3}b^{5/3}c^{2/3}+a^{2/3}b^{2/3}c^{5/3}$ [0]

Now we remember than in the past we saw some inequalities solved proving some "smaller" inequalities that, after being summed up, returned the wanted result.

Let's try now weighted AM-GM on the first two addends of the left member of [0]:

$p_1a^2c+p_2 ab^2 \ge a^{2p_1+p_2}b^{2p_2}c^{p_1}$ [1]

For this to be useful to prove [0], on the exponent of the right member we would like to have $p_1=2p_2$, so that $p_1=2/3,p_2=1/3$. So we recover $b^{2/3}c^{2/3}$, which is what we like. what about the exponent of $a$ ? We have $2p_1+p_2=5/3$. Very good it is going in the right direction. Maybe this is useful ? Let's write [1] with this choice of $p_1$ and $p_2$ :

$\frac{2}{3}a^2c+\frac{1}{3} ab^2 \ge a^{5/3}b^{2/3}c^{2/3}$ [1a]

So now this could be the first piece. Let's try to write other small inequalities for the remaining terms:

$\frac{2}{3}ab^2+\frac{1}{3} bc^2 \ge a^{2/3}b^{5/3}c^{2/3}$ [1b]

$\frac{2}{3}bc^2+\frac{1}{3} a^2c \ge a^{2/3}b^{2/3}c^{5/3}$ [1c]

Summing up [1a],[1b] and [1c] we recover [0] and we are done.

Please write any comments if you find mistakes, especially on the thought process :) .

I am also curious if Muirhead's inequality could help in proving [0] more directly, even if the some on the left is cyclic and not symmetric.

Answer 2

Let $a=e^u$,$b=e^v$,$c=e^w$

The inequality becomes :

Let $u+v+w=0$ then we have :

$$e^{u-v}+e^{v-w}+e^{w-u}\geq e^u+e^v+e^w$$

Or :

$$\sum_{cyc}e^u(e^{-v}-1)\geq 0$$

Or

$$\sum_{cyc}e^u(e^{w+u}-1)\geq 0$$

Now we use weighted jensen's inequality to get :

$$\sum_{cyc}e^u(e^{u+w}-1)\geq (e^u+e^v+e^w)(e^{\frac{(w+u)e^u+(u+v)e^v+(v+w)e^w}{e^u+e^v+e^w}}-1)$$

Remains to show that :

$$\frac{(w+u)e^u+(u+v)e^v+(v+w)e^w}{e^u+e^v+e^w}\geq 0$$

Or :

$$(w+u)e^u+(u+v)e^v+(v+w)e^w\geq 0$$

Wich is not hard and true using Tchebythev's inequality.

Answer 3

We write $a=A^3$, $b=B^3$, $c=C^3$ with $A,B,C>0$ and $ABC=1$. Then $a/b=a^2c=A^6C^3$, and we have to show the homogeneous inequality: $$ A^6C^3 + B^6A^3 + C^6B^3\overset !\ge (A^3+B^3+C^3)A^2B^2C^2\ . $$ For this, use the AM-GM-inequaltity in the form: $$ \frac 13(A^6C^3 + A^6C^3 + B^6A^3) \ge (A^6C^3 \cdot A^6C^3 \cdot B^6A^3) ^{1/3} = A^5B^2C^2 $$ and its cyclic versions.

Answer 4

Because the left sum is cyclic, we can assume $a \le b \le c$ or $c \le b \le a$.

First assume $a \le b \le c$. From $abc = 1$, we can also conclude $a \le 1 \le c$. Then there are $x,y\ge 0$ such that $a = 1/(x + 1)$, $b = (x + 1)/(y + 1)$, $c = y + 1$. It's a bit of algebra, but if you substitute these into the inequality, collect terms, and clear denominators, you'll find it's equivalent to $$ 2(x-y)^2 + P(x, y)\ge 0, $$ where $P$ is a polynomial with only positive coefficients. This is clearly true for all $x, y\ge 0$.

Now apply the same logic in the case $c\le b\le a$. In this case, the polynomial it reduces to is simpler: $$ (x-y)^2(x + y + 2) + x(x+2)y \ge 0 $$ This is also clearly true for all $x,y\ge 0$.

Since the statement is true in both cases, it is true in general.