If a- b- c- d are positive numbers- then ab - cd-2-abcd

As a,b,c,d are positive numbers, So nbsp;ab,cd are also positive nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; Now as AM gt; GM ...

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If a, b, c, d...

Question

If $a, b, c, d$ are positive numbers, then $a b + c d < 2 \sqrt{a b c d}$

True

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False

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Similar questions

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

(a) $B C^{2} + A D^{2} + 2 B C . A D$

(b) $A B^{2} + C D^{2} + 2 A B . C D$

(d) $B C^{2} + A D^{2} + 2 A B . C D$

If $a, b, c, d$ are positive numbers, then $a b + c d < 2 \sqrt{a b c d}$

If $A B$ and $C D$ are a pair of opposite sides of parallelogram $A B C D$ , then

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & a + b + c + d & a b + c d a + b + c + d & 2 (a + b) (c + d) & a b (c + d) + c d (a + b) a b + c d & a b (c + d) + c d (a + b) & 2 a b c d \end{matrix} ∣ ∣ ∣ ∣ ∣

a^{2} + b^{2} = c^{2} + d^{2}

a^{2} + d^{2} - a d = b^{2} + c^{2} + b c

\frac{a b + c d}{a d + b c}

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