Prove that $(x+y)\left(\frac{1}{x}+\frac{4}{y}\right)\ge9,$ if $x,y >0.$

Question

If $x,y>0$, considering $\left(\frac{1}{\sqrt{x}}-\frac{2}{\sqrt{y}}\right)^2$ prove that $(x+y)\left(\frac{1}{x}+\frac{4}{y}\right)\ge9$.

Answer 1

$$(x+y)\left(\frac{1}{x}+\frac{4}{y}\right)=\frac{y}{x}+4\frac{x}{y}+5 =\left(\sqrt{\frac{y}{x}}-2\sqrt{\frac{x}{y}}\right)^2+4+5\ge9$$

Answer 2

Let $$F=(x+y)(1/x+4/y)=1+y/x+4x/y+4=5+(y/x+4x/y)\ge 5+4=9~~~(1)$$ It is by AM-GM that $(A+B) \ge 2\sqrt{AB}$, equality holds if $A=B$. So in (1) equality holds when $y=2x$.

Answer 3

$(x+y)(\frac{1}{x} + \frac{4}{y}) = \frac{y}{x} + 4\frac{x}{y} + 5$

We wish to show that this value is always at least 9. Thus, we wish to show that the minimum of this function is at least 9. Since x,y are both positive we can be assured that this is continuous and differentiable.

Lets define $s=\frac{x}{y}$

We wish to minimize $\frac{1}{s} + 4s + 5 = f(s)$

By simple differentiation, we get $f'(s) = 4 - \frac{1}{s^2}$ which we set equal to zero and solve for s.

Thus clearly $s=\pm\frac{1}{2}$ according to the algebra, but because of how we defined s we know that $s=\frac{1}{2}$ is the only true solution.

Plug this value in for $f(s)$. We get $f(\frac{1}{2})=9$.

We know this is an optimum but is it a minimum?

$f''(s) = 2\frac{1}{s^3}$, and $f''(\frac{1}{2})>0$

We can indeed say that $ \frac{y}{x} + 4\frac{x}{y} + 5 \ge 9$