Prove: Inequality of Sums of Square Roots of Positive Reals

[Saitama]

Homework Statement

Let $a,b,c$ be positive real numbers with sum 3. Prove that
\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca

Homework Equations

AM-GM inequality

The Attempt at a Solution

I don't really know how to start with. We are given $a+b+c=3$.
Also, $2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$ but this doesn't seem helpful here and I don't see how can I apply the AM-GM inequality here. :(

 

[Mandelbroth]

Homework Statement

Let $a,b,c$ be positive real numbers with sum 3. Prove that
\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca

Homework Equations

AM-GM inequality

The Attempt at a Solution

I don't really know how to start with. We are given $a+b+c=3$.
Also, $2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$ but this doesn't seem helpful here and I don't see how can I apply the AM-GM inequality here. :(

I think the general statement of the inequality is $\forall x_i\in\mathbb{R}_+\cup\{0\} \, \frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}$.
I had to think about this, but...let's use your hint.

Evaluate $2x-x^2(3-x^2)$. What does that imply about positive x?
Multiply both sides of the statement you need to prove. See where that gets you.

 

[Saitama]

I think the general statement of the inequality is $\forall x_i\in\mathbb{R}_+\cup\{0\} \, \frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}$.
I had to think about this, but...let's use your hint.

Evaluate $2x-x^2(3-x^2)$. What does that imply about positive x?
Multiply both sides of the statement you need to prove. See where that gets you.

Sorry for the late reply but I figured this out later.

The original inequality (to be proved) can be re-written as
a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \geq 9
This inequality can be easily proved by AM-GM inequality.
a^2+\sqrt{a}+\sqrt{a} \geq 3a
Hence proved.

 

[Mandelbroth]

Sorry for the late reply but I figured this out later.

The original inequality (to be proved) can be re-written as
a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \geq 9
This inequality can be easily proved by AM-GM inequality.
a^2+\sqrt{a}+\sqrt{a} \geq 3a
Hence proved.

I didn't think of that.

$\forall x\geq0, \, 2x−x^2(3−x^2)=(x−1)^2x(x+2)\geq0 \implies 2x \geq x^2(3−x^2)$.

If we multiply both sides of the original inequality by 2, we get $2\sqrt{a}+2\sqrt{b}+2\sqrt{c}\geq 2ab+2bc+2ca = a(b+c)+b(a+c)+c(a+b) = a(3-a)+b(3-b)+c(3-c)$. The proof thus follows as an example of the first inequality by considering the case $x=\sqrt{\alpha}, \, \alpha\in\{a, b, c\}$.