How to prove an Inequality

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 $b=2a$.

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 $a_1, a_2, \ldots, a_n$.

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

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