2017/010) Let $a,\,b,\,c,\,d$ be real with condition that $a+b\sqrt{2}+c\sqrt{3}+2d >= \sqrt{10(a^2+b^2+c^2+d^2)}$

show that $a^2+d^2=b^2+c^2$

Proof:

by Cauchy-Schawartz in equality we have
$(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $\sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)$
from given condition and above we have
$\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
when
$\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)$
So $a = k, b^2= 2k^2,c^2=3k^2,d=2k$ and hence $a^2+d^2=b^2+c^2$

2017/009}Prove that $\sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}\lt 2016$

we have $n^{th}$ term = $\frac{\sqrt{n\cdot (n+1)}}{2n+1}$
$= \frac{\sqrt{n^2+n}}{2n+1}$
$= \frac{\sqrt{n^2+n+\frac{1}{4}-\frac{1}{4}}}{2n+1}$
$=  \frac{\sqrt{(n+\frac{1}{2})^2-\frac{1}{4}}}{2n+1}$
$< \frac{n+\frac{1}{2}}{2n+1}$
$< \frac{1}{2}$
each term is $< \frac{1}{2}$ and there are 4032 terms so sum is less than 2016