Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$

Question

Let $x$ and $y$ be two positive numbers:

Prove that $$ \left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right).$$

I answered this one by squaring the two expressions. And therefore finding the difference after squaring the formulas. I don't even know if it's right but I wanted to find another way to answer this question?

This whole inequality can be gotten from (a-b)^2 > 0. All you have to do is multiple with (x+y) in the end just for special effects ! — Saikat, Feb 14 '16 at 04:16

score 4 · Answer 1 · edited Apr 13 '17 at 12:21

Just notice that the following inequalities are equivalent to each other: $$ \begin{align*} \frac {x+y}{x^2+y^2} &\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)\\ \frac {x+y}{x^2+y^2} &\le \frac12 \cdot \frac{x+y}{xy}\\ \frac 1{x^2+y^2} &\le \frac12 \cdot \frac{1}{xy}\\ 2xy &\le x^2+y^2 \end{align*} $$

The last one is a well-known inequality:

Thank you it was very useful – Manuela NIEVES Feb 08 '16 at 20:57 — Manuela NIEVES, Feb 08 '16 at 20:57

score 1 · Answer 2 · 2016-01-13T20:51:17.470

1

Since $x$ and $y$ are positive, so is $x+y$, and $0\leq(x-y)^2$ since squares are always non-negative. So, $0\leq (x+y)(x-y)^2$, which expands to $0\leq x^3+y^3-xy^2-x^2y$. We can rewrite this to $$(x+y)\cdot xy \leq \tfrac{1}{2}(x(x^2+y^2)+y(x^2+y^2))$$ and dividing by $xy\cdot (x^2+y^2)$ yields $$\frac{(x+y)\cdot xy}{xy(x^2+y^2)}=\frac{x+y}{x^2+y^2} \leq \tfrac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)=\tfrac{1}{2}\frac{x+y}{xy}=\tfrac{1}{2}\frac{x(x^2+y^2)+y(x^2+y^2)}{xy(x^2+y^2)}$$ $$\frac{x+y}{x^2+y^2}\leq\tfrac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)$$ which is the wanted inequality.

Hope this helped!

edited Jan 13 '16 at 20:51

answered Jan 13 '16 at 20:21

How did you pass from the third line to the last one ? I know you divided but can you just please make it clear ? – Manuela NIEVES Jan 13 '16 at 20:48
Thanks for letting me know I was unclear - I edited the answer. Hope it's clear now, but don't hesitate to let me know if you still struggle to understand it! – Jan 13 '16 at 20:52

sinbadh · Answer 3 · 2016-01-13T21:01:32.367

0

Since the inequality is homogenious, you can suppose $x+y=1$. The resulting inequality is the $AM-GM$.

Edit: An inequality $f(x_1,x_2,...,x_n)\ge 0$ is homogenius if there is some $n$ such that for all $f(tx_1,tx_2,...,tx_n)=t^nf(x_1,x_2,...,x_n)$ for all $t>0$. If an inequality is homogenius, you can suppose some condition of normality. For example $x_1+x_2+...+x_n=1$ or $x_1\times x_2\times...\times x_n=1$.

AM-GM is: for $x_1,...,x_n>0$ we have $\frac{x_1+...+x_n}{n}\ge\sqrt[n]{x_1\times...\times x_n}$

edited Jan 13 '16 at 21:01

answered Jan 13 '16 at 20:47

sinbadh

7,521

I don't know about this! We aren't working here with numbers – Manuela NIEVES Jan 13 '16 at 20:49
Mmm i´m using classical results for solving inequalities. Obviously inequalties for numbers – sinbadh Jan 13 '16 at 20:51
What part of my solution don't you understand? – sinbadh Jan 13 '16 at 20:55
I think you meant AM - GM; it stands for Arithmetic Mean - Geometric Mean. – Jan 13 '16 at 20:57
Yeah I think those expressions I didn't quite understood them because of my level of course we didn't have this lesson before – Manuela NIEVES Jan 13 '16 at 20:59

score 0 · Answer 4 · answered Feb 13 '16 at 07:21

An alternative approach:

$\left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right) \Longleftrightarrow 4x^2y + 4xy^2 \leq (x^2 + y^2)(2y + 2x) \Longleftrightarrow x^2y + y^2x \leq x^3 + y^3$

We assume WLOG $x \leq y \Longrightarrow \exists c\in \mathbb{R}_{≥0}$ such that $x+c = y$

Hence the rightmost inequality can be simplified to $0 \leq (x^3 + (x+c)^3) -(x^2(x+c) + x(x+c)^2) = c^2(c+2x)$ which is clear.

Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$

4 Answers4