MathDB

2

Part of 2019 Slovenia Team Selection Test

Problems(2)

Slovenia 2019 TST1 P2

Source: 2019 Slovenia 1st TST Problem 2

2/18/2019
Prove, that for any positive real numbers a,b,ca, b, c who satisfy a2+b2+c2=1a^2+b^2+c^2=1 the following inequality holds. 1a−a+1b−b+1c−c≥2a+2b+2c\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}
TSTinequalities
Slovenia 2019 TST2 P2

Source: 2019 Slovenia 2nd TST Problem 2

2/14/2019
Determine all non-negative real numbers aa, for which f(a)=0f(a)=0 for all functions f:R≥0→R≥0f: \mathbb{R}_{\ge 0}\to \mathbb{R}_{\ge 0} , who satisfy the equation f(f(x)+f(y))=yf(1+yf(x))f(f(x) + f(y)) = yf(1 + yf(x)) for all non-negative real numbers xx and yy.
TSTFunctional Equationsalgebrafunctional equation