39e^x>=1+x for all real `x`
for this i use the convexity of the graph of e^x
draw the graph of both. u can see e^x always lies above 1+x except at the origin where 1+x is tangent to e^x and its the point where equality occurs
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Prove that e^x\geq 1+x\quad\forall \,x\in\mathbb{R}, the set of real numbers (easy part.) Using this result, prove the Arithmetic Mean - Geometric Mean inequality for n non-negative reals (not so easy part).
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3939for the second part i want to wait for others and compare my method.
3taking f(x)=ex-(1+x)
f'(x)=ex -1which is positive so f(x) is an increasing function.
then the equality holds when x=0
11f(x)=ex-x-1
f'(x)=ex-1
Extrema at x=0
f''(x)=ex>0
so at x=0 f(x) has a minima
f(x=0)=0
so f(x)>=0 always
11Well, I know the proof, (coz I'm kaymant sir's student) and to say, the taste of having proved it (with this) is simply gr8...
39yes soumik i absolutely agree with u. i am in really exciting mood to see who is gonna do the second part.
nothing gr88 if nishant sir or prophet sir pick it up.(its cakewalk for them)
341some history: this proof was given by the venerable Hungarian mathematician Polya. Apparently it struck him as he was climbing into a bus!!
39oh!!! my good ness...
u must be a historian rather than a mathematician...
u are world best in mathematics trivia
1After proving the first part,
Take a series, a1,a2,a3,.............,an
Its AM, A= (a1+a2+a3+..........+an)/n
its GM, G= (a1a2a3............an)1/n
Now, for some k>0,
e(ak/A)-1 ≥(ak/A)-1 +1
k=1,
e(a1/A)-1 ≥(a1/A)
k=2,
e(a2/A)-1 ≥(a2/A)
.
.
.
.
k=n,
e(an/A)-1 ≥(an/A)
Multiplying,
e{(a1+a2+a3+..........+an)/A}-n≥(a1a2a3............an)/An
Since, (a1+a2+a3+..........+an)/A =n
e{(a1+a2+a3+..........+an)/A}-n = e0 =1
Therefore,
1≥ (a1a2a3............an)/An
ie., An ≥ (a1a2a3............an)
or, A ≥ (a1a2a3............an)1/n
Hence, A ≥ G
Couldn't have done it, if din do it b4. Knew the beginning!!
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