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State true or false. If x and y are positive real numbers and

Question

State true or false.
If $x$ and $y$ are positive real numbers and $m, n$ are any positive integers, then $\frac{x ^{n} y ^{n}}{( 1 + x ^{2 n} ) ( 1 + y ^{2 m} )} > \frac{1}{4}$

True

False

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Evelyn Discussed

State true or false.

x

and

y

are positive real numbers and

m, n

are any positive integers, then

\frac{x ^{n} y ^{n}}{( 1 + x ^{2 n} ) ( 1 + y ^{2 m} )} > \frac{1}{4}

9 mins ago

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9 mins ago

Text solutionVerified

x

and

y

are positive real numbers and

m

and

n

are positive integers, therefore

\frac{1 + x ^{2 n}}{2} \geq (1 \times x^{2 n})^{1/2}; \frac{1 + y ^{2 m}}{2} \geq (1 \times y^{2 m})^{1/2}

(since for two positive numbers

A . M \geq G . M

)

∴ (\frac{1 + x ^{2 n}}{2}) \geq x^{n} .... (1); (\frac{1 + y ^{2 m}}{2}) \geq y^{m} .... (2)

Multiplying Eqs (1) and (2) we get

\frac{( 1 + x ^{2 n} ) ( 1 + y ^{2 m} )}{4} \geq x^{n} y^{m} \Rightarrow \frac{1}{4} \geq \frac{x ^{n} y ^{m}}{( 1 + x ^{2 n} ) ( 1 + y ^{2 m} )}

hence the statement is false.

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Question Text	State true or false. If $x$ and $y$ are positive real numbers and $m, n$ are any positive integers, then $\frac{x ^{n} y ^{n}}{( 1 + x ^{2 n} ) ( 1 + y ^{2 m} )} > \frac{1}{4}$
Answer Type	Text solution:1
Upvotes	150