Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$

Question

Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$ then find $\frac{b+c}{a}$

So given $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$

So squaring both sides $(\sqrt{a}+9\sqrt{b}+44\sqrt{c})^2=(\sqrt{2018(a+b+c)})^2$

then $a+81b+1936c+18\sqrt{ab}+88\sqrt{ac}+792\sqrt{bc}=2018(a+b+c)$

but it looks much difficult is any other method

Answer 1

By Cauchy Inequality,

$\sqrt{a}+9\sqrt{b}+44\sqrt{c}\le\sqrt{1^2+9^2+44^2}\sqrt{(\sqrt{a})^2+(\sqrt{b})^2+(\sqrt{c})^2}=\sqrt{2018(a+b+c)}$

with equality holds if and only if $\displaystyle \frac{\sqrt{a}}{1}=\frac{\sqrt{b}}{9}=\frac{\sqrt{c}}{44}$, i.e. $$a:b:c=1:81:1936$$