A comment on a simple proof of Cauchy inequality

In mathematics, the inequality

n({{x}_{1}}^{2}+\cdots +{{x}_{n}}^{2})\geq {{({{x}_{1}}+\cdots +{{x}_{n}})}^{2}}

is not hard to prove with quadratic root discriminant.

But the problem is that my computer science roommate want me to give an induction method. So I tried a new method to prove it, in which I used a small trick deserving taking a note.

Let me firstly restate the problem here: it’s known that

{{x}_{1}}+\cdots +{{x}_{n}}=1, prove that {{x}_{1}}^{2}+\cdots+{{x}_{n}}^{2}\geq \frac{1}{n}

The equality holds if and only if

{{x}_{1}}={{x}_{2}}=\cdots ={{x}_{n}}

when

n=1,2,

the inequality obviously holds true.

Assume that when

n=k,

the inequality holds true.

Then when

n=k+1,

if  k+1=2m,

{{x}_{1}}^{2}+\cdots +{{x}_{n}}^{2}=\sum_{i=1}^{2m}{{{x}_{i}}^{2}}=\sum_{i=1}^{m}{({{x}_{i}}^{2}+{{x}_{i+m}}^{2})}\geq \sum_{i=1}^{m}{({{(\frac{{{x}_{i}}+{{x}_{i+m}}}{2})}^{2}}+{{(\frac{{{x}_{i}}+{{x}_{i+m}}}{2})}^{2}})}\geq \frac{1}{2}\sum_{i=1}^{m}{{{({{x}_{i}}+{{x}_{i+m}})}^{2}}}\geq \frac{1}{2}m{{(\frac{\sum_{i=1}^{2m}{{{x}_{i}}}}{m})}^{2}}=\frac{1}{2m}=\frac{1}{n}

if k+1=2m-1,

Therein lies the problem, we cannot break the formula into two parts equally if induction is used.

Instead of proving this inequality directly, we consider another problem which has a subtle link with the  issue we are dealing with. Say, when

{{x}_{1}}+\cdots +{{x}_{2m}}=1+\frac{1}{2m-1}

Prove that

\sum_{i=1}^{2m}{({{x}_{i}}^{2})}\geq 2m{{(\frac{1}{2m-1})}^{2}}.

Actually, by induction method, because

m<2m-1

this statement holds true definitely. Further more, it gives

\sum_{i=1}^{2m-1}{({{x}_{i}}^{2})}\geq \frac{1}{(2m-1)} when \sum_{i=1}^{2m-1}{{{x}_{i}}}=1.

Otherwise, the new problem value would be even smaller.

The inspring point for this method is to add a free variable to cater the old problem, though technical when applying it. And I hope this trick will bring me more stuff.

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One thought on “A comment on a simple proof of Cauchy inequality

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