Not too hard actually

Algebra Level 2

$\large 3(a+b+c)$ If $a,b,c>0$ and $a^2+b^2+c^2+abc=4$ , find the maximum value of the expression above

The answer is 9.

2 solutions

P C
Mar 14, 2016

Call the expression P, assume that $a\geq b\geq c$ . In three numbers $a,b,c$ there'll be two numbers larger than or equal to 1, in this case they are $a$ and $b$ . So now we have $(a-1)(b-1)\geq 0$ $\Leftrightarrow ab\geq a+b-1$ Based on the condition we have $4-c^2=(2-c)(2+c)=a^2+b^2+abc\geq 2ab+abc=ab(2+c)$ , therefore $2-c\geq ab$ $\Rightarrow 2-c\geq a+b-1$ $\Leftrightarrow 3(a+b+c)\leq 9$ The equality holds when $a=b=c=1$

This Dirichlet's theorem ? I don't see the relevance. Can you elaborate on this? Thanks.

I think you meant to change all the $a,b,c$ 's to $x,y,z$ 's.

Pi Han Goh - 5 years, 3 months ago

I'm sorry, my mistake, what I used to solve is the pigeonhole pinciple

P C - 5 years, 3 months ago

Why must this be true?

In three numbers $a,b,c$ there'll be two numbers larger than or equal to 1, we assume that they are $a$ and $b$ .

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – maybe because of the statement a > b> c

Daniel Shamsudin - 5 years, 3 months ago

a=b=c :v .Nice

Son Nguyen - 5 years, 3 months ago

Brilliant :)

Muhammad Arafat - 5 years, 2 months ago

Jase Jason
Mar 15, 2016

I think logically and all I figured out was that if x, y, or Z would be bigger than 1, it would not satisfy the equation. Any smaller, and it would not satisfy the equation either. I tried 1 as a value for all, and apparently, the question maker made my brain explode with rage after 30 minutes wasted thinking

Not too hard actually

The answer is 9.

2 solutions

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