The minimal

Algebra Level 4

Find the square of the minimum value of 9 + x 2 + ( 4 − x ) 2 + ( 3 − y ) 2 + 4 + y 2 \sqrt{9+x^{2}}+\sqrt{(4-x)^{2}+(3-y)^{2}}+\sqrt{4+y^{2}} as x x and y y range over all real numbers.


The answer is 72.

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Joel Tan by Joel Tan , 21, Singapore

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4 solutions

Harry Ray
May 28, 2016

Relevant wiki: Classical Inequalities

I came across this problem in the geometry section, so I thought I'd provide a geometric solution.

The sum of square root terms suggests three straight-line segments. Consider the following diagram: The first line segment has length x 2 + 9 \sqrt{x^2 + 9} , the second has length y 2 + 4 \sqrt{y^2 + 4} , and the final segment has length: ( 6 − ( x + 2 ) ) 2 + ( 6 − ( y + 3 ) ) 2 = ( 4 − x ) 2 + ( 3 − y ) 2 \sqrt{(6 - (x+2))^2 + (6 - (y+3))^2} = \sqrt{(4-x)^2 + (3-y)^2} That is, the quantity we want to minimize is simply the sum of the lengths of these three line segments. This is minimized when they form a straight line from the origin to ( 6 , 6 ) (6,6) , which has a length of 6 2 6\sqrt{2} hence the answer is 72 \boxed{72} .

6 Helpful 0 Interesting 3 Brilliant 0 Confused

Brilliant! This is a very simple and elegant solution. Thanks for sharing this approach, Harry :)

Pranshu Gaba - 5 years ago
Kartik Sharma
Dec 16, 2014

9 + x 2 + ( 4 − x ) 2 + ( 3 − y ) 2 + 4 + y 2 \sqrt{9 + {x}^{2}} + \sqrt{{(4-x)}^{2} + {(3-y)}^{2}} + \sqrt{4 + {y}^{2}}

Using Cauchy-Schwarz in Titu form,

( 1 2 + 1 2 ) ( 3 2 + x 2 ) ≥ ( 3 + x ) 2 ({1}^{2} + {1}^{2})({3}^{2} + {x}^{2}) \geq {(3 + x)}^{2}

9 + x 2 ≥ ( 3 + x ) 2 2 9 + {x}^{2} \geq \frac{{(3 + x)}^{2}}{2}

9 + x 2 ≥ x + 3 2 \sqrt{9 + {x}^{2}} \geq \frac{x + 3}{\sqrt{2}}

Similarly,

( 4 − x ) 2 + ( 3 − y ) 2 ≥ 7 − x − y 2 \sqrt{{(4-x)}^{2} + {(3-y)}^{2}} \geq \frac{7-x-y}{\sqrt{2}}

4 + y 2 ≥ y + 2 2 \sqrt{4 + {y}^{2}} \geq \frac{y + 2}{\sqrt{2}}

Adding all these inequalities,

9 + x 2 + ( 4 − x ) 2 + ( 3 − y ) 2 + 4 + y 2 ≥ x + 3 2 + 7 − x − y 2 + y + 2 2 \sqrt{9 + {x}^{2}} + \sqrt{{(4-x)}^{2} + {(3-y)}^{2}} + \sqrt{4 + {y}^{2}} \geq \frac{x + 3}{\sqrt{2}} + \frac{7-x-y}{\sqrt{2}} + \frac{y + 2}{\sqrt{2}}

9 + x 2 + ( 4 − x ) 2 + ( 3 − y ) 2 + 4 + y 2 ≥ 12 2 \sqrt{9 + {x}^{2}} + \sqrt{{(4-x)}^{2} + {(3-y)}^{2}} + \sqrt{4 + {y}^{2}} \geq \frac{12}{\sqrt{2}}

≥ 6 2 \geq 6\sqrt{2}

Hence, the answer is 72 \boxed{72}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Same , But just for sake of variety , I used the fact that RMS value is always greater than the average value. And I think all of us Proved it in our schools , In while studying A l t e r n a t i n g C u r r e n t Alternating Current chapter (Otherwise we can also proove it by using 2-D co-ordinate geometry , i.e by using s t r a i g h t L i n e s straight Lines ). Hence :

R m s ≥ A v g a 2 + b 2 2 ≥ a + b 2 a 2 + b 2 ≥ a + b 2 \displaystyle{Rms\ge Avg\\ \sqrt { \cfrac { { a }^{ 2 }+b^{ 2 } }{ 2 } } \ge \cfrac { a+b }{ 2 } \\ \sqrt { { a }^{ 2 }+b^{ 2 } } \ge \cfrac { a+b }{ \sqrt { 2 } } }

So Applying This concept on this question , we get 9 + x 2 + ( 4 − x ) 2 + ( 3 − y ) 2 + 4 + y 2 ≥ 12 2 \displaystyle{\sqrt { 9+{ x }^{ 2 } } +\sqrt { { (4-x) }^{ 2 }+{ (3-y) }^{ 2 } } +\sqrt { 4+{ y }^{ 2 } } \ge \cfrac { 12 }{ \sqrt { 2 } } }

Equality Occurs , Means Minimum is achieve when values becomes equal on which we are doing Inspection . Means when a = b a=b .

Hence x = 3 x=3 , y = 2 y=2 .

Deepanshu Gupta - 6 years, 3 months ago

can u explain it a bit further??

Nithin Nithu - 6 years, 1 month ago
Arturo Presa
May 21, 2017

Let us represent the given expression by f ( x , y ) , f(x, y), and then, f ( x , y ) = ∣ 3 i + x j ∣ + ∣ ( 4 − x ) i + ( 3 − y ) j ∣ + ∣ 2 i + y j ∣ . f(x, y)=|3i+xj|+|(4-x)i+(3-y)j|+|2i+yj|. Using the triangle inequality and the inequality a 2 + b 2 ≥ ( a + b ) 2 2 , a^2+b^2\geq \frac{(a+b)^2}{2}, we obtain that f ( x , y ) ≥ ∣ ( 3 + 4 − x + 2 ) i + ( x + y + 3 − y ) j ∣ = ( 9 − x ) 2 + ( x + 3 ) 2 ≥ ( 9 − x + x + 3 ) 2 2 = 72 . f(x,y)\geq |(3+4-x+2)i+(x+y+3-y)j|=\sqrt{(9-x)^2+(x+3)^2}\geq \sqrt{\frac{(9-x+x+3)^2}{2}}=\sqrt{72}.

Besides that, all the quantities involved in these two consecutive inequalities become equal when x = 3 , x=3, and y = 2. y=2. Then, the minimum value of f ( x , y ) f(x, y) for any positive values of x x and y y is 72 . \sqrt{72}. Then the answer to the question is 72 . \boxed{72}. So, in my opinion, the condition that the values of x x and y y are positive is not necessary.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ashutosh Sharma
Feb 12, 2018

its easy to do in complex number system,

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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