Don't make mistake

Algebra Level 4

Given that $a,b$ and $c$ are real numbers satisfying $\frac{1}{a^2+8}+\frac{1}{4b^2+8}+\frac{1}{9c^2+8}=\frac{1}{3}.$ Find the product of the minimum and maximum value of the expression below. $a+2b+3c$

Inspiration .

The answer is -9.

2 solutions

Akshay Sharma
Feb 9, 2016

Using Jensen inequality on $f(x)=(\frac{1}{x^2+8})$ with $x_i =a,2b,3c$ we get , $(a+2b+3c)^2\leq 9$ ,Hence product of maximum and minimum value of $a+2b+3c$ is $-9$

Can you further explain it, cause I'm not good at Jensen's Inequality

P C - 5 years, 4 months ago

This may help you.

Akshay Sharma - 5 years, 4 months ago

thanks, i understand now

P C - 5 years, 4 months ago

Further I have edited the solution.

Akshay Sharma - 5 years, 4 months ago

Please explain, how can you use jensen's inequality even though the function is not purely convex or purely concave...?

Manuel Kahayon - 5 years, 4 months ago

Harish Ghunawat
Feb 9, 2016

Let $a^2+8=4b^2+8=9c^2+8=1/9$ so sum will become $1/3$ so $a=\pm1;b=\pm2\;c=\pm3$ so maximum value is $3$ and min is $-3$ then product is $-9$

Why must $a^2+8=4b^2+8=9c^2+8$ ?

Pi Han Goh - 5 years, 4 months ago

@Pi Han Goh Do you have a reason Pi?

Akshay Krishna - 2 years, 5 months ago

Can you tell why it happens to be max and min when they are equal?

Akshay Krishna - 2 years, 5 months ago

Don't make mistake

The answer is -9.

2 solutions

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