8(abcd+1)>(a+1)(b+1)(c+1)(d+1) prove it

if a,b,c,d,all are greater than 1

Note by Aditya Kumar
7 years, 12 months ago

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Comments

What have you tried? What do you know?

As a hint : If x,y>1x, y >1 , then (x1)(y1)>0 (x-1)(y-1) > 0 so xy+1>x+y xy + 1 > x + y . Use this many times.

Calvin Lin Staff - 7 years, 12 months ago

this question was from the beginning section of the book in inequalities so it does not require any prerequisities so i tried so solve this without am gm inequalitiy or any other but cannot solve it

Aditya Kumar - 7 years, 11 months ago

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As a hint : If x,y>1x,y>1, then (x1)(y1)>0(x-1)(y-1)>0 so xy+1>x+yxy+1>x+y.

This looks very similar to the inequality, especially on the LHS. Use this many times.

Calvin Lin Staff - 7 years, 11 months ago

Just expand it , and try to put in square form , its very simple

Shivang Jindal - 7 years, 11 months ago

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I think, its from Challenge and thrill . right?

Shivang Jindal - 7 years, 11 months ago

I agree with just expand it. I'm not so sure what you mean by square form, though there is a way to write it as the sum of several positive (non-negative) terms.

Calvin Lin Staff - 7 years, 11 months ago

thanks i have now solved it

Aditya Kumar - 7 years, 11 months ago

Can someone explain it once more??...

Tanmay Dokania - 3 years, 8 months ago

Try to substitute x=a1,y=b1,z=c1,w=d1x=a-1, y=b-1, z=c-1, w=d-1.

Shourya Pandey - 7 years, 11 months ago

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Don't you mean x=a+1,y=b+1x=a+1,y=b+1, etcetera?

Tim Vermeulen - 7 years, 11 months ago

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No

Shourya Pandey - 7 years, 11 months ago

Try doing it, it becomes very easy

Shourya Pandey - 7 years, 11 months ago
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