The result f ( 3 x 1 + x 2 + x 3 ) ≤ 3 f ( x 1 ) + f ( x 2 ) + f ( x 3 ) follows from f ( 4 x 1 + x 2 + x 3 + x 4 ) ≤ 4 f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 ) , if x 4 in the latter is replaced by

We can test one-by-one and determine that 3 x 1 + x 2 + x 3 is the correct replacement.

The result f ( 3 x 1 + x 2 + x 3 ) ≤ 3 f ( x 1 ) + f ( x 2 ) + f ( x | TestHub

Passage / Comprehension

Let $f$ be a function satisfying $f (\frac{x _{1} + x _{2}}{2}) \leq \frac{f ( x _{1} ) + f ( x _{2} )}{2}$ , for all $x_{1}, x_{2}$ in its domain. If we draw the graph of a continuous function satisfying this inequality, we will notice that the chord joining any two points on the curve will always be above the portion of the curve between the two points. We can also prove that if the given inequality is true, then a similar result follows for four values $x_{1}, x_{2}, x_{3}, x_{4}$ . Indeed:

$f (\frac{x _{1} + x _{2} + x _{3} + x _{4}}{4}) = f (\frac{\frac{x _{1} + x _{2}}{2} + \frac{x _{3} + x _{4}}{2}}{2}) \leq \frac{f ( \frac{x _{1} + x _{2}}{2} ) + f ( \frac{x _{3} + x _{4}}{2} )}{2}$

$\leq \frac{\frac{f ( x _{1} ) + f ( x _{2} )}{2} + \frac{f ( x _{3} ) + f ( x _{4} )}{2}}{2} \leq \frac{f ( x _{1} ) + f ( x _{2} ) + f ( x _{3} ) + f ( x _{4} )}{4}$

The result $f (\frac{x _{1} + x _{2} + x _{3}}{3}) \leq \frac{f ( x _{1} ) + f ( x _{2} ) + f ( x _{3} )}{3}$ follows from $f (\frac{x _{1} + x _{2} + x _{3} + x _{4}}{4}) \leq \frac{f ( x _{1} ) + f ( x _{2} ) + f ( x _{3} ) + f ( x _{4} )}{4}$ , if $x_{4}$ in the latter is replaced by

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Solution:

We can test one-by-one and determine that $\frac{x _{1} + x _{2} + x _{3}}{3}$ is the correct replacement.

Mathematics - Application of Derivative Question with Solution | TestHub

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Doubts & Discussion