58. National mathematical olympiad Final round

First solution. Let $d_1,\dots ,d_n$ be real numbers and $$f(x)=\sum _{i,j=1}^n\frac{d_i{d}_j}{c_i+c_j}{x}^{c_i+c_j},x>0.$$ Since $x{f}^{\prime }(x)=({\sum }_{i=1}^n{d}_i{x}^{c_i}{)}^2\ge 0$, it follows that $f(x)\ge f(0+)=0$. In particular, $$f(1)=\sum _{i,j=1}^n\frac{d_i{d}_j}{c_i+c_j}\ge 0.$$ Then for any real number $t$, $$0\le \sum _{i,j=1}^n\frac{(a_{i}t+b_i)(a_{j}t+b_j)}{c_i+c_j}=A{t}^2+2Ct+B,$$ where $$A=\sum _{i,j=1}^n\frac{a_i{a}_j}{c_i+c_j},B=\sum _{i,j=1}^n\frac{b_i{b}_j}{c_i+c_j},C=\sum _{i,j=1}^n\frac{a_i{b}_j}{c_i+c_j}.$$ This implies that $AB\ge C^2$.

Second solution. Set $f(x)={\sum }_{i=1}^n{a}_i{x}^{c_i-1/2}$ and $g(x)={\sum }_{i=1}^n{b}_i{x}^{c_i-1/2}$. The given inequality follows by the Cauchy-Schwarz inequality: $${\int }_0^1{f}^2(x)dx{\int }_0^1{g}^2(x)dx\ge {\left({\int }_0^{1}f(x)g(x)dx\right)}^2.$$