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It obviously holds when

n = 1, 2

if nimm is true, then:

$$ 2 (\ sqrt {MIMO 1}-1)\ le\ sum_ {KMUL1} ^ m\ frac {(KMUE 1)!} {KMB!} $$

$$ 2 (\ sqrt {m}-1)\ le\ sum_ {KMUL1} ^ {MMUL1}\ frac {(KMUL1)!} {KFEN!} $$

$$ 2 (\ sqrt {mmur1}-1)\ le\ sum_ {klut1} ^ {mmur2}\ frac {(kmur1)!} {kink!} $$

when n=m+1:

$$ Left = 2 (\ sqrt {masked 2}-1) $$

$$ Right side =\ sum_ {KFL1} ^ {MFLY1}\ frac {(KMUL1)!!} {KFEN!} =\ sum_ {KFL1} ^ m\ frac {(KMUL1)!!} {KFLY1!} +\ frac {MUL1!} {(KMUL1)!!} $$

so just prove the following:

$$ \ sum_ {KMUL1} ^ m\ frac {(KMU1)!!} {KMUL1!} +\ frac {MY1!!} {(MUBL1)!!}-2 (\ sqrt {MUBE2}-1)\ ge 0 $$

.

the next step is to find a way to prove this inequality. But put

$$ \ sum_ {KMUL1} ^ m\ frac {(KMU1)!} {KFEN!} $$

replace directly with:

$$ 2 (\ sqrt {masked 1}-1) $$

No (as I've done before), it can cause you to zoom too much. I haven't thought of the method of proof yet.

in addition

$$ \ frac {masked!} {(masked 1)!!} $$

can be written as

$$ \ frac {(mmur2)!!} {(mmel1)!!} *\ frac {m} {mmel1} $$

this may be used in the derivation process.

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discuss separately according to the parity of $n $. Only even numbers are discussed below. If $n $is odd, it can be proved by similar method.

if $n$ is an even number $n = 2mcm:

$$ \ array { & &\ sum_ {Kwon 1} ^ n\ frac {(KMU1)!!} {KFN!}\ hfill\\ & &\ sum_ {Kenz1} ^ m\ frac {(2k-1)!!} {(2k)!!} +\ sum_ {KFL1} ^ m\ frac {(2k-2)!!} {(2k-1)!}\ hfill &\ text {(separate summation of odd and even terms)}\ hfill\\ & = &\ sum_ {Kwon 1} ^ m\ frac {(KMel 1ppb 2)!} {k!} +\ sum_ {KLY 1} ^ m\ frac {(KMY 1)!} {2 (KMAE 1max 2)!}\ hfill &\ text {(the denominator continuously removes the common factor 2)}\ hfill\\ & = &\ left (\ frac {2 (mpound 1 right 2)!} {m!}-1\ right) +\ left (\ frac {m!} {(m *)} {(m *)} {(m *)}-1\ right)\ hfill &\ text {(for inductive proof, see below)}\ hfill\\ &\ ge& 2\ sqrt {\ frac {2 (mmur1Acer 2)!} {(mmel1Acer 2)!}}-2\ hfill &\ text {(basic inequality)}\ hfill\\ & = & 2\ sqrt {2 (masked 1hip 2)}-2\ hfill\ hfill\\ & = & 2 (\ sqrt {naugh1}-1)\ hfill } $$

In the second step, you still use the symbol $! $to denote the factorial of a semi-integer, such as $(5ax 2)! = (5ax 2) (3max 2) (1max 2) $. The conclusion of the third step can be concluded by induction. Such as proving

$$ \ sum_ {Kwon 1} ^ m\ frac {(KMel 1A 2)!} {k!} =\ frac {2 (m!}-1)!} {m!}-1 $$

as long as you verify that the equation holds for $mm2 $, and

$$ \ array { & &\ frac {2 (masks 1 hfill 2)!} {m!}-1 +\ frac {(masks 1-1 machetes 2)!} {(masks 1)!}\ hfill\\ & = &\ frac {(masked 1 + 2)! (2 (masked 1) + 1)} {(masked 1)!}-1\ hfill\\ & = &\ frac {2 ((masked 1) + 1 hfill 2)!} {(masked 1)!}-1\ hfill } $$