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Algebra

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8 Feb 2012 12:02:56 IST

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inequality.

Mathematics

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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Comments (3)

rishabh

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Joined: 13 Oct 2011

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8 Feb 2012 16:52:35 IST

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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)

jagdish singh

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10 Feb 2012 20:21:13 IST

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$\hspace{-16}$Let $\mathbf{S=a+b+c}$ and $\mathbf{Q=ab+bc+ca}$ and $\mathbf{a,b,c>0}$\\\\ Let $\mathbf{x=a+b+1\;\;,y=b+c+1\;\; z=c+a+1}$\\\\ Now Using $\mathbf{A.M\geq G.M}$ Inequality, We Get\\\\ $\mathbf{S=a+b+c\geq 3.\left(abc\right)^{\frac{1}{3}}=3}$ bcz $\mathbf{abc=1}$ \\\\ So $\mathbf{S\geq 3}$ Similarly\\\\ $\mathbf{Q=ab+bc+ca\geq 3.\left(a^2b^2c^2\right)^{\frac{1}{3}}=3}$ bcz $\mathbf{abc=1}$\\\\ So $\mathbf{Q\geq 3}$\\\\ So $\mathbf{S.(Q-2)\geq 3\Leftrightarrow S^3-S^2.S+SQ-1-2S-2\geq 0}$\\\\ $\mathbf{(S-a).(S-b).(S-c)-(2S+3)+1\geq 0}$\\\\ $\mathbf{(b+c).(c+a).(a+b)-(2a+2b+2c+3)+1\geq 0}$\\\\ $\mathbf{(y-1)(z-1)(x-1)-(x+y+z)+1\geq 0}$\\\\ $\mathbf{xyz\geq xy+yz+zx}$\\\\ $\mathbf{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 1}$$

hemang

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11 Feb 2012 08:42:25 IST

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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.

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