# Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.

Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

Here's my idea:

$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$

And I'm stuck here.

I need to prove that:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$

In the first case using Cauchy-Schwarz Inequality I prove that:

$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$

$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$

Now I need to prove that:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$

$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$

I need I don't know how to continue.

In the second case I tried proving:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and

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

Using Cauchy-Schwarz Inequality I proved:

$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$

$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$

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

But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$

P.S

My initial idea, which is proving:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$

maybe isn't the right way to prove this inequality.

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You could try doing something with infinitesimal calculus to show that the minimum of $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2)$ occurs only when there is some other specific relationship between $a$, $b$, and $c$. –  AJMansfield Mar 20 '13 at 23:19
I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. For help with formatting in the future, please see this meta question. P.S. Thank you very much for showing your work. Too few people do that, but it helps raise/maintain the standard of the site. :) –  anorton Mar 20 '13 at 23:19

I will use the following lemma (the proof below):

$$2x \geq x^2(3-x^2)\ \ \ \ \text{ for any }\ x \geq 0. \tag{\clubsuit}$$

Start by multiplying our inequality by two

$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq 2ab +2bc +2ca, \tag{\spadesuit}$$

and observe that

$$2ab + 2bc + 2ca = a(b+c) + b(c+a) + c(b+c) = a(3-a) + b(3-b) + c(3-c)$$

and thus $(\spadesuit)$ is equivalent to

$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq a(3-a) + b(3-b) + c(3-c)$$

which can be obtained by summing up three applications of $(\clubsuit)$ for $x$ equal to $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ respectively:

\begin{align} 2\sqrt{a} &\geq a(3-a), \\ 2\sqrt{b} &\geq b(3-b), \\ 2\sqrt{c} &\geq c(3-c). \\ \end{align}

$$\tag*{\square}$$

The lemma

$$2x \geq x^2(3-x^2) \tag{\clubsuit}$$

is true for any $x \geq 0$ (and also any $x \leq -2$) because

$$2x - x^2(3-x^2) = (x-1)^2x(x+2)$$

is a polynomial with roots at $0$ and $-2$, a double root at $1$ and a positive coefficient at the largest degree, $x^4$.

$\hspace{60pt}$

I hope this helps ;-)

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How do you rearrange? Explain that step, because $$(\sqrt{a} + \sqrt{a})^2 + (2\sqrt{b})^2 + 4c = 4(a+b+c)$$ How do we know that $$a^2 + b^2 + c^2 + 2\sqrt{a} + 2\sqrt{b} + 2\sqrt{c} + 3 \geq (\sqrt{a} + \sqrt{a})^2 + (2\sqrt{b})^2 + 4c$$ ? –  Stefan4024 Mar 21 '13 at 0:04
@Stefan4024 That step was wrong, check out the new proof. –  dtldarek Mar 21 '13 at 8:02
Good work!!! Thanks you a lot. It was so good explained that even a kid from first grade will get it. :D –  Stefan4024 Mar 21 '13 at 10:24

From the given inequality $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ observe that $$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$$ We can rewrite the original inequality as
$$a^2+2\sqrt{a}+ b^2+2\sqrt{b}+ c^2+2\sqrt{c}\ge9$$ since $(a+b+c)=3$. Using AM-GM set the LHS up as follows: $$a^2+\sqrt{a}+\sqrt{a}\ge3\sqrt[3]{a^2 \sqrt{a}\sqrt{a}}=3a$$ $$b^2+\sqrt{b}+\sqrt{b}\ge3\sqrt[3]{b^2 \sqrt{b}\sqrt{b}}=3b$$ $$c^2+\sqrt{c}+\sqrt{c}\ge3\sqrt[3]{c^2 \sqrt{c}\sqrt{c}}=3c$$ Adding the three inequalities yields $$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 3(a+b+c) =9$$ with equality if an only if $a$=$b$=$c$=$1$.

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Hint:

What lower bound does AM-GM give you when you consider $a^2 + \sqrt{a} + \sqrt{a}$?

Your hope that $\sum \sqrt{a} \ge \sum a = 3$ is false, by using Cauchy Schwarz: $9 = 3(\sum a) \ge (\sum \sqrt{a})^2$. In fact, when $a+b+c = 3$, we have $$\sum a^2 \ge \sum a = 3 \ge \sum \sqrt{a}$$ all by Cauchy-Schwarz, so your hope to split the inequality up is thwarted. This also signals us that we should try to "mix" $a^2$ and $\sqrt{a}$ together in some way, hence the hint.

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## Hint

From the given equation:

\begin{align} \sqrt{a} + \sqrt{b} + \sqrt{c} &\ge ab + bc + ca \\ \dfrac{a}{\sqrt{a}} + \dfrac{b}{\sqrt{b}} + \dfrac{c}{\sqrt{c}} &\ge ab + bc + ca \\ \dfrac{a \sqrt{bc} + b \sqrt{ac} + c \sqrt{ab}}{\sqrt{abc}} &\ge ab + bc + ca \end{align}

Now, take $( \sqrt{a} + \sqrt{b} + \sqrt{c} )$. We have

\begin{align} ( \sqrt{a} + \sqrt{b} + \sqrt{c} )^2 &= a + b + c + 2 \left( \sqrt{ab} + \sqrt{bc} + \sqrt{ca} \right) \\ &= 3 + 2 \left( \sqrt{ab} + \sqrt{bc} + \sqrt{ca} \right) \\ & \geq 0 \end{align}

We already have \begin{align} ( a + b +c )^2 &= a^2 + b^2 + c^2 + 2 \left( ab + bc + ca \right) \\ &= 9 \\ & \geq 0 \end{align}

Thus the following relation can be arrived at:

\begin{align} \left( ( \sqrt{a} + \sqrt{b} + \sqrt{c} )^2 - 2 ( \sqrt{ab} + \sqrt{bc} + \sqrt{ca} \right)^2 = a^2 + b^2 + c^2 + 2 \left( ab + bc + ca \right) \end{align}

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