# Prove $(a + b)^2 \geq 4ab$

Prove $(a + b)^2 \geq 4ab$

What direction should I take with this proof? Can I use induction here? or is there a better method?

I tried a few manipulations, but couldn't seem to find a form that proved it for all $x$.

One such manipulation resulted in:

$a^2 + b^2 \geq 2ab$

which seems close to the triangle inequality. Can I use this somehow?

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Note that you can't use induction, since you want to prove that this is true for all real numbers $a$ and $b$. Induction is mainly useful for proving statements about integers. –  Zev Chonoles Mar 26 '12 at 4:40
You should include what a and b are. I assume they are real numbers, otherwise this statement is not necessarily true. –  vsz Mar 26 '12 at 5:13
Use the substitution: $$a=r\cos\theta , b=r\sin\theta$$ –  Sawarnik Feb 28 at 7:07

You are on the right track; but try subtracting $4ab$ from both sides of the original inequality, instead of just $2ab$. Do you recognize what you get on the left? Finally, remember that $x^2\geq0$ for any real number $x$.
I would have $(a - b)^2$ and it is true that this term is always $\geq 0$ for $a$ and $b \in \mathbb{R}$. Seems too easy : ) –  stariz77 Mar 26 '12 at 4:47
You're pretty close. Set it up as a series of if and only if statements. $$(a+b)^2\geq 4ab \Leftrightarrow a^2+2ab+b^2\geq 4ab \Leftrightarrow a^2-2ab+b^2\geq 0 \Leftrightarrow (a-b)^2\geq 0$$ and the last statement is clearly always true.