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hemang (1555)

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given that a, b and c are positive real numbers.

and abc = 1.

prove that 1/(a + b + 1) + 1/(b + c + 1) + 1/(a+c+1) < or equal to 1.

my solution is a bit long. wanted to see something interesting with it.


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There are 3 types of people in the world. Those who can count and those who can't . :-)

    
rishabh (0)

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dividing denominator by abc1/{1/bc + 1/ac +abc} + 1/{1/ac + 1/ab +abc} 1/{1/bc + 1/ab +abc}it is equal toabc{1/abc+2 + 1/abc+2 + 1/abc+2}maximum value of abc=1then the expression equals to1/3 +1/3 +1/3=1(max value is 1)
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jagdish singh (1031)

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Jagdish Singh.(Pantnagar)
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hemang (1555)

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awesome as always jagdish bhaiya!!

here is my take on it.

i let x = a + b.

y = b + c.

z = c + a.

we get,

1/(y+1) + 1(z+1) < o equal to x/(x+1).

or, on croos multiplication,

x + y + z + 2 < or equal to xyz.

2(a+b+c) + 2 < or equal to (a+b)(b+c)(c+a) = a^2b + b^2a + c^2a + a^2c + b^2c + c^2b + 2abc.

we cancel out 2abc and 2.

then we add 3 to both the sides.

a^2b + a^2c + 1 > or equal to 3a by AM - GM.

similalrly for all and thenn add.

2(a+b+c) + 3 < or equal to 3(a+b+c).

or a + b + c > or equal to 3.

which is evidient from am - gm since abc = 1.

 

 


mathematics is the food of my soul.
There are 3 types of people in the world. Those who can count and those who can't . :-)

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