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I'm beginner with proofs and I got the follow exercise:

Prove the inequality $$(a + b)\Bigl(\frac{1}{a} + \frac{4}{b}\Bigr) \ge 9$$ when $a > 0$ and $b > 0$. Determine when the equality occurs.

I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ?

Thanks in advance.

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We only need $ab>0$ in your inequality. – blindman Oct 5 '12 at 3:57
up vote 6 down vote accepted

The first thing to do is simplify the expression on the lefthand side of the inequality.

$$(a + b)\left(\frac{1}{a} + \frac{4}{b}\right)=\frac{(a+b)(4a+b)}{ab}=\frac{4a^2+5ab+b^2}{ab}=\frac{4a}b+5+\frac{b}a\;.$$

Now notice that the resulting expression contains both $\frac{a}b$ and $\frac{b}a$; this is an indication that it might simplify matters to introduce a new quantity, $x=\frac{a}b$, and rewrite the inequality as $$4x+5+\frac1x\ge 9\;,$$ or $$4x+\frac1x\ge 4\;.\tag{0}$$

The natural thing to do now is to multiply through by $x$ to get rid of the fraction, but be careful: since this is an inequality, the sign of $x$ matters. If $x\ge 0$ we get $4x^2+1\ge 4x$, or $4x^2-4x+1\ge 0$, but if $x<0$ we get $4x^2+1\le 4x$, or $4x^2-4x+1\le 0$. In either case, though, we recognize that $4x^2-4x+1=(2x-1)^2$, so either $$x\ge 0\quad\text{and}\quad(2x-1)^2\ge 0\tag{1}$$ or $$x<0\quad\text{and}\quad(2x-1)^2\le 0\;.\tag{2}$$

Now $(2)$ is impossible: $(2x-1)^2\le 0$ if and only if $(2x-1)^2=0$, in which case $2x=1$, $x=\frac12$, and $x\not<0$. Thus, any solution must come from $(1)$: $x>0$, and $(2x-1)^2\ge 0$. If $2x\ne 1$, then $2x-1\ne0$, so $(2x-1)^2>0$, and we have a solution. If $2x=1$, then $x=\frac12>0$ and $(2x-1)^2=0\ge0$, and again we have a solution. In short, every positive $x$ is a solution, no negative $x$ is a solution, and $x$ can’t be $0$. (Why not?) We could actually have discovered this just by looking more closely at $(0)$, but one won’t always have so nice an inequality as that.

What does this mean in terms of $a$ and $b$? Recall that $x=\dfrac{a}b$; thus, $x>0$ if and only if $\dfrac{a}b>0$, which is true if and only if $a$ and $b$ have the same algebraic sign: both are positive, or both are negative. Since we were told that both are positive, we know that the inequality holds for all $a$ and $b$ in the given domain.

Finally, we have equality in $(0)$ if and only if $4x^4-4x+1=0$, or $(2x-1)^2=0$, i.e., if and only if $x=\frac12$. Since $x=\frac{a}b$, that’s equivalent to $\frac{a}b=\frac12$, or $b=2a$.

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Your argument is interesting and native. It will be helpful for the author of this question. – blindman Oct 5 '12 at 4:00
Thank you, it was definitely helpful. – aajjbb Oct 12 '12 at 0:45
@aajjbb:You’re welcome; I’m glad it helped. – Brian M. Scott Oct 12 '12 at 6:55

Recalling the Bunhiacopski inequality $$ (a_1b_1+a_2b_2+\ldots+a_nb_n)^2 \leq (a_1^2+a_2^2+\ldots+a_n^2)(b_1^2+b_2^2+\ldots+b_n^2).$$ Applying the above inequality with $n=2$, $a_1=\sqrt{a}, a_2=\sqrt {b}$, $b_1=1/\sqrt{a}, b_2=2\sqrt{b}$ we get $$ 9=(1+2)^2=(a_1b_1+a_2b_2)^2\leq(a_1^2+a_2^2)(b_1^2+b_2^2)=(a+b)\left(\frac{1}{a}+\frac{4}{b}\right). $$ By using the similar arguments we deduce a generalized inequality $$ (a_1+a_2+\ldots+a_n)\left(\frac{1}{a_1}+\frac{2^2}{a_2}+\ldots+\frac{n^2}{a_n}\right)\geq \left(\frac{n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{4} $$ for all positive real numbers $a_1, a_2, \ldots, a_n$.

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@aajjbb: I gave the solution and the generalization of your problem. – blindman Oct 5 '12 at 3:47

If you are familiar with Cauchy-Schwarz, this is an immediate conesquence:

$$ \left( \sqrt{a}\frac{1}{\sqrt{a}}+ \sqrt{b} \frac{2}{\sqrt{2}} \right)^2 \leq (a+b)(\frac{1}{a}+\frac{4}{b} )$$

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Simplify the inequality and you will get $\frac{4a}{b}+\frac{b}{a}\geq 4$. Now let $x=\frac{a}{b}$ and it becomes $4x+\frac{1}{x}\geq 4 \Leftrightarrow 4x+\frac{1}{x}-4\geq 0 \Leftrightarrow \left(2\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 \geq 0$ which is always true. The equality holds when $2\sqrt{x}=\frac{1}{\sqrt{x}}$ or $x=\frac{1}{2} \Leftrightarrow b=2a$

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We can play with the inequality: let's suppose it was true for now, and perform a series of reversible steps and see what the inequality would imply. So we begin by multiplying everything out, which gives:

$1 + 4 + \dfrac{b}{a} + \dfrac{4a}{b} \geq 9 \Leftrightarrow \dfrac{b}{a} + \dfrac{4a}{b} \geq 4$. (1)

Now, when working with inequalities there are some common results that you use; one of these is the AM-GM inequality which states:

$x + y \geq 2\sqrt{xy}$,

where $x$ and $y$ are nonnegative. (Square both sides and collect the terms to one side to see why this is true).

Letting $x = b/a$ and $y = 4a/b$ yields that

$\dfrac{b}{a} + \dfrac{4a}{b} \geq 2\sqrt{\dfrac{b}{a}\cdot \dfrac{4a}{b}} = 2\sqrt{4} = 4$,

which is what we wanted to show in line (1).

Since we can reverse all of our steps (we expanded and subtracted 5; to reverse this add 5 and then factor), we have proven the original inequality. (Note: a formal solution would start with this last line and work "upwards.")

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You can calculate as following:

$$\begin{align*} (a+b)(\frac{1}{a}+\frac{4}{b})&=\frac{b}{a}+\frac{4a}{b}+5\\ &=(\frac{b}{a}+\frac{4a}{b}-4)+9\\ &=\frac{4a^{2}-4ab+b^{2}}{ab}+9\\ &=\frac{(2a-b)^{2}}{ab}+9\\ &\geq9 \end{align*}$$

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Another one...
$ (a+b)({1 \over b} + {4 \over a}) \ge 9 \qquad \qquad $
This requires that also $a,b\ne 0$ and thus $a,b \gt 0$ as from the problem-definition

Then multiplication with ab does not introduce inconsistencies and we can do
$ (a+b)({1 \over a} + {4 \over b}) \ge 9 \qquad \qquad // *ab $
$ (a+b)(b + 4a) \ge 9ab \qquad \qquad // $ expand
$ 4a^2+5ab+b^2 \ge 9ab \qquad \qquad // - 9ab $
$ 4a^2-4ab+b^2 \ge 0 \qquad \qquad $

(1) $ \qquad (2a-b)^2 \ge 0 \qquad \qquad $ This is always true

(2) $ \qquad $ The case of equality is now obvious...

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Simple algebra yields $$ 1+\frac{4a}{b}+\frac{b}{a} +4=5+\frac{4a}{b}+\frac{b}{a} $$ If $a \geq b $ then the inequality is obvious. The second case is a bit trickier. The inequality then reduces to proving that (writing $0<\frac{a}{b}=\epsilon<1$) $$ 4 \epsilon^2-4 \epsilon+1 \geq 0 $$ Which holds for all $\epsilon$ and the equality is at $2a=b$.

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You made a mistake on your argument. The inequality $$4\epsilon^2-4\epsilon+1\geq 0$$ holds for all $\epsilon$ since $4\epsilon^2-4\epsilon+1=(2\epsilon-1)^2\geq0$. – blindman Oct 5 '12 at 3:50
thanks, corrected. – Alex Oct 6 '12 at 23:28

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