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# Inequality 15(Kvant)

Problem:

Prove that for any real numbers $\displaystyle a_1,a_2,...,a_n$ the following inequality holds:

$\displaystyle \left(\sum^{n}_{i=1}a_{i}\right)^{2}\leq \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}$.

Solution:

Observe that

$\displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}=\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt$.

But $\displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}$ can be considered as a constant. So,

$\displaystyle \sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt=\int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt$.

Notice now that

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt=\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt$.

So, the inequality reduces to

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\sum^{n}_{i=1}a_{i}\right)^{2}$.

Now, using Cauchy-Schwartz inequality for integrals, we get that

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}$.

But, we must now observe that

$\displaystyle \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}=\left(\sum^{n}_{i=1}a_{i}\right)^{2}$,

which comes to the conclusion, Q.E.D.