How to prove that there exist $a$ and $b$ in [0,1] such that $\left| f\left( a \right) +g\left( b \right) -ab \right| \ge \frac { 1 }{ 4 } $? Let $f$ and $g$ be any real-valued functions defined on [0,1],Prove that there exist $a$ and $b$ in [0,1] 
such that  
$$\left| f\left( a \right) +g\left( b \right) -ab \right| \ge \frac { 1 }{ 4 }  $$


*

*I don't know how can I start with this question. So , I need some hints which can help me to approach this problem .


Let us prove it from contradiction.
Suppose that there exist such $f$ and $g$, that 
$$
\left| f\left( a \right) +g\left( b \right) -ab \right| < \frac { 1 }{ 4 }
$$
Consider endpoint of $a$ and $b$, which leads to the following system of inequalities:
$$
-\frac{1}{4}<f(0)+g(0)-0<\frac{1}{4}\\
-\frac{1}{4}<f(1)+g(0)-0<\frac{1}{4}\\
-\frac{1}{4}<f(0)+g(1)-0<\frac{1}{4}\\
-\frac{1}{4}<f(1)+g(1)-1<\frac{1}{4}
$$
Add 2nd and 3rd inequalities, subtract from them 1st:
$$
-\frac{3}{4}<f(1)+g(1)<\frac{3}{4}
$$
While 4th implies:
$$
\frac{3}{4}<f(1)+g(1)<\frac{5}{4}
$$
Hence, no $f$ and $g$ satisfying initial equation exist. QED.
