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Let $x,y\in\mathbb{R^n}$ and $K>0$ prove that $$ \frac{2K||x-y||_2^2}{ (1+K||x||_2^2)(1+K||y||_2^2) } <2 $$

What I've achieved so far is:

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$$ =\frac{8K||x||_2^2}{(1+K||y||_2^2)^2} $$

I've tried several things like doing a case-distinction on $K$, or using the bounds $(1+Kx^2)^2\geq (1+Kx^2)\geq Kx^2$ But I never get below the bound $2$.

ndrizza
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2 Answers2

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$2K \| x - y \|^{2} \leq 2(1+K \|x\|^{2} )(1 + K \|y\| ^{2})$

If and only if

$K \|x-y\| ^{2} \leq 1 + K (\|x\|^{2} + \|y\|^{2}) + K^{2} \|x\|^{2} \|y\|^{2}$

We have that $ K\|x-y\|^{2} \leq K(\|x\| + \|y\|)^{2} = K(\|x\|^{2} + 2\|x\| \|y\| + \|y\|^{2})$

And $K(\|x\| + \|y\|)^{2} = K\|x\|^{2} + 2K\|x\| \|y\| + K\|y\|^{2} \leq 1 + K (\|x\|^{2} + \|y\|^{2}) + K^{2} \|x\|^{2} \|y\|^{2}$

if and only if $0 \leq 1-2K\|x\| \|y\| + K^{2} \|x\|^{2} \|y\|^{2} = (1-K\|x\| \|y\|)^{2}$

So, we are done.

Note that if $n = 2, x = (1,0), y = (0,1)$ and $K = 1$

Then $ \frac{2K \| x - y \|^{2}}{(1+K \|x\|^{2} )(1 + K \|y\| ^{2})} = 2$ so is not true "<", change "$\leq$" instead

ZAF
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Strict inequality fails as pointed out by ZAF.

It is a good idea to get rid of all the vectors and the norms by trying to prove that $\frac {2K(a+b)^{2}} { {(1+Ka^{2})}{(1+Kb^{2})}} \leq 2$ for positive numbers $a$ and $b$. Once you prove this the given inequality holds since $\|x-y\| \leq \|x\|+\|y\|$.

Some simple algebraic manipulation reduces this inequality to $(1-Kab)^{2} \geq 0$ which is true.

Kavi Rama Murthy
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