 # a,b,x,y, are negative numbers a^5 + b^5<= 1 x^5 + y^5 <= 1 prove that a^2x^3 + b^2y^3<= 1 Tammy Todd 2021-02-25 Answered
a,b,x,y, are negative numbers
${a}^{5}+{b}^{5}\le 1$
${x}^{5}+{y}^{5}\le 1$
prove that ${a}^{2}{x}^{3}+{b}^{2}{y}^{3}\le 1$
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The numbers ${a}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}^{2}$ will always be positive as these are squared numbers.
The numbers ${x}^{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}^{3}$ will always be negative numbers as it is the cube of negative numbers and the cube of negative numbers is always negative numbers.
${x}^{3}<0$ and${y}^{3}<0$ if x,y<0
Therefore,
${a}^{2}{x}^{3}<0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}^{2}{y}^{3}<0$ if x,y<0 .
The sum of negative numbers is also negative . Therefore,
${a}^{2}{x}^{3}+{b}^{2}{y}^{3}<0$ (1)
Now if a=x, b=y. Then,
${a}^{2}{x}^{3}+{b}^{2}{y}^{3}={a}^{2}\left({a}^{3}\right)+{b}^{2}\left({b}^{3}\right)$
$={a}^{5}+{b}^{5}$
$\le 1$
if $a=x,b=y,{a}^{2}{x}^{3}+{b}^{2}{y}^{3}\le 1$ (2)
Now combine the results of equation (1) and (2).
Hence, ${a}^{2}{x}^{3}+{b}^{2}{y}^{3}\le 1$ has been proved.