a,b,x,y, are negative numbers a5+b5≤1 x5+y5≤1 prove that a2x3+b2y3≤1

The numbers a2andb2 will always be positive as these are squared numbers. The numbers x3andy3 will always be negative numbers as it is the cube of negative numbers and the cube of negative numbers is always negative numbers. x3&lt;0 andy3&lt;0 if x,y&lt;0 Therefore, a2x3&lt;0andb2y3&lt;0 if x,y&lt;0 . The sum of negative numbers is also negative . Therefore, a2x3+b2y3&lt;0 (1) Now if a=x, b=y. Then, a2x3+b2y3=a2(a3)+b2(b3) =a5+b5 ≤1 if a=x,b=y,a2x3+b2y3≤1 (2) Now combine the results of equation (1) and (2). Hence, a2x3+b2y3≤1 has been proved.

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High School
Algebra
Pre-Algebra
Negative numbers and coordinate plane

Tammy Todd

Answered question

2021-02-25

a,b,x,y, are negative numbers

a^{5} + b^{5} \leq 1

x^{5} + y^{5} \leq 1

prove that

a^{2} x^{3} + b^{2} y^{3} \leq 1

Answer & Explanation

Roosevelt Houghton

Skilled2021-02-26Added 106 answers

The numbers $a^{2} and b^{2}$ will always be positive as these are squared numbers.
The numbers $x^{3} and y^{3}$ 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,
$a^{2} x^{3} < 0 and b^{2} y^{3} < 0$ 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,
$a^{2} x^{3} + b^{2} y^{3} = a^{2} (a^{3}) + b^{2} (b^{3})$
$= a^{5} + b^{5}$
$\leq 1$
if $a = x, b = y, a^{2} x^{3} + b^{2} y^{3} \leq 1$ (2)
Now combine the results of equation (1) and (2).
Hence, $a^{2} x^{3} + b^{2} y^{3} \leq 1$ has been proved.

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