Inequality in Equality

Algebra Level 4

For positive real numbers a , b , c , d a,b,c,d how many solutions does the following equation have?

7 ( a + b + c + d ) ( a b c + a b d + b c d + a c d ) = 100 a b c d 7(a+b+c+d)(abc+abd+bcd+acd) = 100abcd


The answer is 0.

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Fahim Shahriar Shakkhor by Fahim Shahriar Shakkhor , 24, Bangladesh

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2 solutions

Eddie The Head
Mar 7, 2014

Adjusting both sides we get, ( a + b + c + d ) ( 1 / a + 1 / b + 1 / c + 1 / d ) = 100 7 = 14.282828......... (a+b+c+d)(1/a+1/b+1/c+1/d) = \frac{100}{7} = 14.282828.........

By AM-GM inequality for positive reals a , b , c , d a,b,c,d , we have, ( a + b + c + d ) ( 1 / a + 1 / b + 1 / c + 1 / d ) ≥ 16 (a+b+c+d)(1/a+1/b+1/c+1/d) \geq 16

Hence there are no solutions...

22 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice!

Calvin Lin Staff - 7 years, 3 months ago

Really cool!

Anik Mandal - 6 years, 1 month ago
Nicholas Fortino
Mar 29, 2014

Using AM-GM we get a+b+c+d >= 4(abcd)^(1/4) and (abc+bcd+acd+abd) >= 4(a^3b^3c^3d^3)^(1/4) . 7(a+b+c+d)(abc+bcd+acd+abd) >= 7 16 abcd=112 abcd . since the minimum value of 7(a+b+c+d)(abc+bcd+acd+abd) > 100abcd hence there are no solution

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