# 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 '14 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 $x$.

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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
It does - but it's just the awesomeness of math at work :) –  Zev Chonoles Mar 26 '12 at 4:50

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.

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short proof $$(a+b)^2=4ab+(a-b)^2$$

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Here is an alternative proof, which is circuitous but perhaps enjoyable as a curiosity:

Fact: For any fixed $P > 0$, the rectangle of perimeter $P$ with maximal area is a square.

Consequence: Given a $2a\times 2b$ rectangle, it has area $4ab$ and perimeter $4a + 4b$. But the square of the same perimeter has side-length $a+b$, hence area $(a+b)^2$.

The Fact above now provides the desired inequality in terms of square and rectangle areas:

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

Fact Proof (for completeness): Let us denote one side of the rectangle by $x$ and the other by $y$. Then the perimeter is $P = 2x + 2y$, so that $2y = P - 2x$, whence $y = P/2 - x$.

Now the area is $xy = x(P/2 - x) = Px/2 - x^2$. To maximize the last expression, observe that it is a downward facing parabola; so its maximum occurs at the vertex, i.e., $P/4$.

Thus, $x = P/4 = y$ as claimed.

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First note that $\forall a,b\in\mathbb{R}$ $$(a-b)^{2}\ge 0.$$ $$\Rightarrow$$ $$a^{2}+b^{2}\ge2ab$$$$\Rightarrow$$

$$a^{2}+b^{2}+2ab\ge 4ab$$ $$\Rightarrow$$ $$(a+b)^{2}\ge 4ab.$$ So, a good method for solving these types of problems is to assume that the inequality is true and see where it brings you, then work backwards. Let me know if you need any clarifications!

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