Question

Se consideră funcția f:R->R,f(x)=
$\sqrt[3]{x + 1}$
Demonstrați că:
f(8)+f(26)>f(124)
​

Answer 1

 

$\displaystyle\bf\\\sqrt[\b3]{\bf9}>\sqrt[\b3]{\bf8}~deoarece~9>8\\\\f(8)+f(26)=\sqrt[\b3]{\bf8+1}+\sqrt[\b3]{\bf26+1}=\sqrt[\b3]{\bf9}+\sqrt[\b3]{\bf27}\\\\f(7)+f(26)=\sqrt[\b3]{\bf7+1}+\sqrt[\b3]{\bf26+1}=\sqrt[\b3]{\bf8}+\sqrt[\b3]{\bf27}=2+3=\boxed{\bf5}\\\\f(124)=\sqrt[\b3]{\bf124+1}=\sqrt[\b3]{\bf125}=\boxed{\bf5}\\\\\\\implies~f(8)+f(26)>f(7)+f(26)=f(124)=5\\\\\implies~\boxed{\bf~f(8)+f(26)>f(124)}$