Constrained minimum with inequalities

Question

Show that for all real positive $x$ and $y$ such that $x+y=1$, the following holds: $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}.$$ I know this can probably be easily solved with Lagrange multipliers, but I am more interested in proving it using inequalities. I think it could be profitable to use AM-HM or something similar but I am not sure about how to proceed. How can I do it? Thanks.

Answer 1

This problem can be rewritten as:

$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq\frac{5}{2}$$

We can use inequality between square and arithmetic mean:

$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq \frac{(x+\frac{1}{x})+(y+\frac{1}{y})}{2}=\frac{1+\frac{1}{x}+\frac{1}{y}}{2}$$

We only have to show, that $$\frac{1}{x}+\frac{1}{y} \geq 4$$

But this comes from $x+y=1$ and inequality between arithmetic and harmonic mean:

$$\frac{1}{2}=\frac{x+y}{2}\geq\frac{2}{\frac{1}{x}+\frac{1}{y}} \Rightarrow \frac{1}{x}+\frac{1}{y} \geq 4$$

Answer 2

Or we can write the left-hand side as $$(x^2+y^2)(1+\frac{1}{x^2y^2})\geq \frac{17}{2}$$ and this is with $$x^2+y^2=1-2xy$$ $$1-2xy+\frac{1}{x^2y^2}-\frac{2}{xy}\geq \frac{17}{2}$$ now let $$xy=t$$ then we get $$-1/2\,{\frac { \left( 4\,t-1 \right) \left( {t}^{2}+4\,t+2 \right) }{ {t}^{2}}} \geq 0$$