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Algebra Level 3

$\large \dfrac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$

If $a,b$ and $c$ are positive real numbers, find the minimum value of the expression above.

The answer is 125.

3 solutions

Paolo Pró
Oct 22, 2015

Factorise one a, b and c from each factor in the numerator. Cancel out abc and the expression becomes (a+3+1/a)(b+3+1/b)(c+3+1/c). Now, let f(x)=x+3+1/x be a function (similar to the factors above). Taking the first derivative, we get f'(x)=1-1/x^2. To minimise, this must be 0 and if we solve for x, the only possible value is 1 (a,b and c are positive). Taking the second derivative, we verify it's a minimum: f''(1)=2/1^3>0. Thus, each factor has been minimised to its smallest possible value by taking all variables equal to 1. Replacing in the expression, we get 125

Sahil Sharma
Oct 3, 2015

From denominator, place 'a' under (a2+3a+1) & similarly do the same for 'b' & 'c'..now separately look at a terms... It will become [ a+1/a +3] & as we know for any positive real no x--> x+1/x >=2...so the minimum value of this bracket will be (2+3)=5....similarly other brackets will also give us minimum value of 5 each... Thus overall minimum value will be 5x5x5=125.

Talha Ali
Oct 3, 2015

The smallest Positive integer is "1". So Let a=1, b=1, c=1. now equation is [{1^2+3(1)+1}{(1)^2+3(1)+1}{1^2+3(1)+1}]/(1)(1)(1) =125

your solution is absolutely correct but you have to mention why you are taking all the 3 digits equal to 1 as asked by #Kushagra Sahni

Atul Shivam - 5 years, 8 months ago

yes your solution is absolutely correct

Atul Shivam - 5 years, 8 months ago

No, You must show that the minimum value of this expression is 125 without taking any particular values of x,y and z. Furthermore, it is nowhere written that x,y,z are integers.

Kushagra Sahni - 5 years, 8 months ago

Exactly one does not simply assign numbers

Mardokay Mosazghi - 5 years, 6 months ago

Bri

The answer is 125.

3 solutions

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