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I had some homework and wasn't really sure how to do it. I just started proofs and this is really new to me.

Question is Suppose $x, y$ are positive real numbers. Show that $$(x^2 - y^2)\left(\frac1y - \frac1x\right) \geq 0.$$

Suppose $x, y$ are positive real numbers. Show that $$\sqrt x +\sqrt y \leq \sqrt{\frac{x^2}{y}} + \sqrt{\frac{y^2}{x}}.$$

I would appreciate any help also if you can explain your answers so I can understand what went on.

Thanks in advance.

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    $\begingroup$ You really should format your question using latex. $\endgroup$
    – ADA
    May 18, 2017 at 3:05
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    $\begingroup$ For (1), just rewrite the LHS as $\dfrac{(x-y)^2(x+y)}{xy}$. Since $x,y$ are positive and squares of reals are non-negative, we arrive at the conclusion that it is $\geq 0$ and equality holds iff $x=y$. $\endgroup$ May 18, 2017 at 3:07
  • $\begingroup$ The first part has already been asked(twice) for answers see here math.stackexchange.com/questions/2278510/… $\endgroup$
    – clark
    May 18, 2017 at 3:19
  • $\begingroup$ math.stackexchange.com/questions/2278738/… $\endgroup$ May 18, 2017 at 3:41

4 Answers 4

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If $x> y$ both factors are positive. If $x< y$ both factors are negative. If $x=y$ both factors are zero.

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$(x^2 - y^2)(\frac 1y - \frac 1x) = \frac {x^2}{y} - x - \frac {y^2}{x} + y = \frac {(x^2 - 2xy + y^2)(x+y)}{xy} = \frac {(x-y)^2(x+y)}{xy}$

$(x-y)^2\ge 0$ for all real $x,y$ and $x+y> 0$ and $xy > 0$ if $x,y>0$

$\frac {(x-y)^2}{xy}\ge 0$

$\sqrt x + \sqrt y \le \sqrt \frac {x^2}{y} + \sqrt \frac {y^2}{x}$

since both sides must be greater that 0, we can square both sides without worry.

$x + y + 2\sqrt {xy} \le \frac {x^2}{y} + \frac {y^2}{x} + 2\sqrt {xy}\\ x + y \le \frac {x^2}{y} + \frac {y^2}{x}\\ (x+y)xy \le x^3 + y^3\\ (x+y)xy \le (x + y)(x^2 - xy + y^2)\\ xy \le x^2 - xy + y^2\\ 0 \le (x- y)^2$

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For the second problem, notice that if $x,y$ are positive real numbers, then $\sqrt x, \sqrt y$ also are positive real numbers. Let $u=\sqrt x$ and $v=\sqrt y$; then the statement to prove is $$u+v \leq \frac{u^2}{v} + \frac{v^2}{u}.$$

Now if you can show that the inequality above is equivalent to $(u^2-v^2)\left(\frac1v - \frac1u\right) \geq 0$ (which it is), you can use the fact that you just proved in the first problem.

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For the first inequality, you could approach it in two ways:

  • Show that if $x,y > 0$, then $x^2 - y^2 < 0 \iff \frac{1}{y} - \frac{1}{x} < 0$, i.e. both terms in the inequality are greater than or equal to $0$, or both are negative
  • Show that if $x\geq y > 0$, then $x^2 - y^2 \geq 0 $ and $\frac{1}{y} - \frac{1}{x} \geq 0$, so their product is $\geq 0$, and symmetrically show that if $y > x>0$ then $x^2 - y^2 < 0 $ and $\frac{1}{y} - \frac{1}{x} < 0$, so their product is still positive.

For the second inequality, as others have mentioned, you should start by squaring both sides to eliminate those pesky square roots. Then see if you can show that this inequality is equivalent (by arithmetic manipulation) to the first one.

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