Inequality (1 Viewer)

Ephi · Apr 14, 2016

√(p^2+q^2) + √(r^2+s^2) >= √(p^2+q^2+r^2+s^2)

for all real p, q, r, s.

True or false.

InteGrand · Apr 14, 2016

Ephi said:
√(p^2+q^2) + √(r^2+s^2) >= √(p^2+q^2+r^2+s^2)

for all real p, q, r, s.

True or false.

True

kawaiipotato · Apr 14, 2016

Ephi said:
√(p^2+q^2) + √(r^2+s^2) >= √(p^2+q^2+r^2+s^2)

for all real p, q, r, s.

True or false.

Triangle inequality

Paradoxica · Apr 14, 2016

InteGrand said:
True

By Inspection.

leehuan · Apr 15, 2016

Basically what kawaiipotato said. To visualise it:

$$The triangle inequality states that $|x|+|y| \ge |x+y|$\\So just let $x=\sqrt{a},y=\sqrt{b}\\\sqrt{a}+\sqrt{b}\ge\sqrt{ab}$

InteGrand · Apr 15, 2016

leehuan said:
Basically what kawaiipotato said. To visualise it:

$$The triangle inequality states that $|x|+|y| \ge |x+y|$\\So just let $x=\sqrt{a},y=\sqrt{b}\\\sqrt{a}+\sqrt{b}\ge\sqrt{ab}$

$\noindent Doing this shows that $\left | \sqrt{a} \right | + \left |\sqrt{b} \right | \geq \left | \sqrt{a} +\sqrt{b}\right |$, which is just equivalent to an equality as both sides are equal, so this doesn't prove the claim.$

$\noindent The claim is easily proved though by showing $LHS^2 \geq RHS^2$. Then since the LHS and RHS are both non-negative, it follows that $LHS\geq RHS$.$

$\noindent In fact, for any concave function $f$ with domain $\mathbb{R}_{\geq 0}$ and with $f(0)\geq 0$, we have that $f$ is subadditive, i.e. $f(a+b) \leq f(a) + f(b)$ (a reversal of this inequality applies for convex $f$). The inequality in the original post is just a special case of this where the function $f$ is the square root function.$

leehuan · Apr 15, 2016

Should've wrote |x|=sqrt(a) but I got lazy

Edit: Hold on nvm I just realised I break the RHS by doing so. Yeah alright my bad.

KingOfActing · Apr 15, 2016

InteGrand said:
$\noindent Doing this shows that $\left | \sqrt{a} \right | + \left |\sqrt{b} \right | \geq \left | \sqrt{a} +\sqrt{b}\right |$, which is just equivalent to an equality as both sides are equal, so this doesn't prove the claim.$

$\noindent The claim is easily proved though by showing $LHS^2 \geq RHS^2$. Then since the LHS and RHS are both non-negative, it follows that $LHS\geq RHS$.$

$\noindent In fact, for any concave function $f$ with domain $\mathbb{R}_{\geq 0}$ and with $f(0)\geq 0$, we have that $f$ is subadditive, i.e. $f(a+b) \leq f(a) + f(b)$ (a reversal of this inequality applies for convex $f$). The inequality in the original post is just a special case of this where the function $f$ is the square root function.$

If we're assuming the triangle inequality as a given, just let a = p + qi, b = b = r + si

|a| + |b| >= |a+b| by triangle inequality

then just writing the definition of the modulus of a complex number gives the required inequality

InteGrand · Apr 15, 2016

KingOfActing said:
If we're assuming the triangle inequality as a given, just let a = p + qi, b = b = r + si

|a| + |b| >= |a+b| by triangle inequality

then just writing the definition of the modulus of a complex number gives the required inequality

$\noindent Well if we wanted to do it like this, we'd get $a+b = \left(p+r\right)+\left(q+s\right)i$, so $|a+b| = \sqrt{\left(p+r\right)^2 + \left(q+s\right)^2}= \sqrt{p^2 + q^2 + r^2 + s^2 + 2pr + 2qs}$. So the triangle inequality would give us $\sqrt{p^2 + q^2}+ \sqrt{r^2 + s^2} \geq \sqrt{p^2 + q^2 + r^2 + s^2 + 2pr + 2qs}$, which doesn't immediately imply the desired inequality. But this is easily done by first assuming that $p,q,r,s$ are non-negative, in which case the inequality \emph{is} implied. Then since the inequality has been proven for non-negative $p,q,r,s$, it is also true for the case where some of them are negative, since all the terms present are squares. So the inequality holds for all real $p,q,r,s$.$

Ephi · Apr 15, 2016

Spectacular answers. Appreciated.

Inequality (1 Viewer)

Ephi

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