Oh, those are vector!

Algebra Level 4

a 2 + a b + 2 b 2 + b 2 + b c + 2 c 2 + c 2 + a c + 2 a 2 \sqrt{a^2+ab+2b^2}+\sqrt{b^2+bc+2c^2}+\sqrt{c^2+ac+2a^2}

Given that a , b a,b and c c are real numbers satisfying a + b + c ≥ 1007 a+b+c\geq1007 , find the minimum value of the expression above.

Bonus : Try to solve this problem using vectors.


The answer is 2014.

View wiki
Son Nguyen by Son Nguyen , 20, Vietnam

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Son Nguyen
Nov 14, 2015

Ok a 2 + a b + 2 b 2 = ( a + 1 2 b ) 2 + 7 4 b 2 \sqrt{a^2+ab+2b^2}=\sqrt{(a+\frac{1}{2}b)^2+\frac{7}{4}b^2} b 2 + b c + 2 c 2 = ( b + 1 2 c ) 2 + 7 4 a 2 \sqrt{b^2+bc+2c^2}=\sqrt{(b+\frac{1}{2}c)^2+\frac{7}{4}a^2} c 2 + a c + 2 c 2 = ( c + 1 2 a ) 2 + 7 4 c 2 \sqrt{c^2+ac+2c^2}=\sqrt{(c+\frac{1}{2}a)^2+\frac{7}{4}c^2} Use ∣ → a ∣ + ∣ → b ∣ + ∣ → c ∣ ≥ ∣ → a + → b + → c ∣ \left | \underset{a}{\rightarrow} \right |+\left | \underset{b}{\rightarrow} \right |+\left | \underset{c}{\rightarrow} \right |\geq \left | \underset{a}{\rightarrow}+\underset{b}{\rightarrow}+\underset{c}{\rightarrow} \right | → a = ( ( a + 1 2 b ) 2 ; 7 4 b 2 ) \underset{a}{\rightarrow}=((a+\frac{1}{2}b)^2;\frac{7}{4}b^2) Also like that with → b , → c \underset{b}{\rightarrow},\underset{c}{\rightarrow} Put A= a 2 + a b + b 2 + . . . + c 2 + a c + 2 a 2 \sqrt{a^2+ab+b^2}+...+\sqrt{c^2+ac+2a^2} A ≥ ∣ a 2 + b 2 + c 2 + a b + b c + a c + 1 4 ( a 2 + b 2 + c 2 ) ; 7 4 ( a 2 + b 2 + c 2 ) ∣ A\geq \left | a^2+b^2+c^2+ab+bc+ac+\frac{1}{4}(a^2+b^2+c^2);\frac{7}{4}(a^2+b^2+c^2) \right | With a + b + c ≥ 1007 a+b+c\geq 1007 We have A ≥ 2 ( a + b + c ) = 2 × 1007 = 2014 A\geq 2(a+b+c)=2\times 1007=2014

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Otto Bretscher
Nov 14, 2015

Applying Jensen's inequality to the convex function f ( x , y ) = x 2 + x y + 2 y 2 f(x,y)=\sqrt{x^2+xy+2y^2} , we see that the given sum is f ( a , b ) + f ( b , c ) + f ( c , a ) ≥ 3 f ( ( a + b ) + ( b , c ) + ( c , a ) 3 ) = 2 ( a + b + c ) ≥ 2014 f(a,b)+f(b,c)+f(c,a)\geq 3 f\left(\frac{(a+b)+(b,c)+(c,a)}{3}\right)=2(a+b+c)\geq \boxed{2014}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

To verify that we have a convex function in 2 variables, we have to calculate the Hessian (matrix of second order partial derivatives) and show that is is positive semidefinite.

I used this inequality

∣ → a ∣ + ∣ → b ∣ + ∣ → c ∣ ≥ ∣ → a + → b + → c ∣ \left | \underset{a}{\rightarrow} \right |+\left | \underset{b}{\rightarrow} \right |+\left | \underset{c}{\rightarrow} \right |\geq \left | \underset{a}{\rightarrow}+\underset{b}{\rightarrow}+\underset{c}{\rightarrow} \right |

Son Nguyen - 5 years, 7 months ago

Reply to Challenge Master: Alternatively, for those who are not familiar with the Hessian, we could show that f f is convex on any line in the plane.... this boils down to a simple Calculus I exercise.

Otto Bretscher - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...