S T = x + 2 + y + 5 + z + 1 0 = x + 1 + y + 1 + z + 1
If the above is true for x , y , z > 0 , what is the minimum value of S 2 − T 2 ?
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Beautiful!!
Yeah I solved it with the obvious and quicker one but this way is really amazing.
I did the same as you. Love this solution!
Excellent solution!
I think Cauchy-Schwarz is way more straightforward:
S 2 − T 2 = ( S + T ) ( S − T ) = [ ( x + 2 − x + 1 ) + ( y + 5 − y + 1 ) + ( z + 1 0 − z + 1 ) ] [ ( x + 2 + x + 1 ) + ( y + 5 + y + 1 ) + ( z + 1 0 + z + 1 ) ] = [ ( x + 2 − x + 1 ) + ( y + 5 − y + 1 ) + ( z + 1 0 − z + 1 ) ] [ x + 2 − x + 1 1 + y + 5 − y + 1 4 + z + 1 0 − z + 1 9 ] ≥ ( 1 + 2 + 3 ) 2 = 3 6
The equality holds when
⎩ ⎪ ⎨ ⎪ ⎧ x + 2 − x + 1 = 1 y + 5 − y + 1 = 2 z + 1 0 − z + 1 = 3 Which is accurately x = y = z = 0 .
but x=y=z=0 is not allowed, [math]x,y,z>0[/math]
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(x,y,z)=(3,15,35) is allowed, too.
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But it is not trivial right? How to find it?
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@Kelvin Hong – View Yang Daniel's solution, it clearly shows that the minimum occurs at x + 1 : y + 1 : z + 1 = 1 : 2 : 3
x = y = z = 0 is NOT a solution. In that case S 2 − T 2 ≈ 3 7 . 4 1 0 9 6 . Perhaps you meant x = 0 , y = 3 , z = 8 ?
With regards to the x > 0 comment: if you take x infinitesimally greater than 0, you will be able to get S 2 − T 2 arbitrarily close to 36. Close enough to be able to convincingly state that the minimum value of 36 is indeed possible, even though it is reached at the border of the domain.
As X X mentions in response to the comment below infinitely many solutions for S 2 − T 2 = 3 6 do exist, also completely inside the domain.
sqrt(z+10) - sqrt(z+1) is not equal to 3 for any integere z.
Sqrt(15+10) - sqrt(15+1) = 1
sqrt(0+10) - sqrt(0+1) = sqrt(10) - 1 is less than 3.
By cauchy inequality S 2 − T 2 = ( S + T ) ( S − T ) = ( x + 2 + y + 5 + z + 1 0 + x + 1 + y + 1 + z + 1 ) ( x + 2 + y + 5 + z + 1 0 − x + 1 − y + 1 − z + 1 ) = ( x + 2 + x + 1 + y + 5 + y + 1 + z + 1 0 + z + 1 ) ( x + 2 + x + 1 1 + y + 5 + y + 1 4 + z + 1 0 + z + 1 9 ) ≧ ( 1 + 2 + 3 ) 2 = 3 6
Hi, there are typos at third last line.
Sir, your reasoning is wrong, base on your calculation the equality required negative value of x , y , z to hold.
thank you, the mistake has been corrected
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NO! Now the equality doesn't hold also.
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Why the equality doesn't hold ?you know there is almost no difference between your method and mine
对x,y,z>0,取不到最小值36吧?要x=y=z=-1才取到36,请问原题出自哪呢?
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(x,y,z)=(3,15,35) is allowed, too.
Since S 2 − T 2 = ( S + T ) ( S − T ) , the product rule gives us d x d ( S 2 − T 2 ) = 2 S d x d S − 2 T d x d T .
At minimum this must be 0, so x + 2 S = x + 1 T T 2 S 2 = x + 1 x + 2 T 2 S 2 − 1 = x + 1 1 Doing this for y and z as well we get T 2 S 2 − 1 = x + 1 1 = y + 1 4 = z + 1 9 Introducing a constant k such that k 1 = T 2 S 2 − 1 and using the original definition T = 1 + x + 1 + y + 1 + z , we can express T at minimum in terms of k: T = k + 4 k + 9 k = 6 k T 2 = 3 6 k
And using the definition for k, we find S 2 = ( k 1 + 1 ) T 2 = 3 6 + 3 6 k In the difference k does not appear, so we have our answer: S 2 − T 2 = 3 6
Recall that by QM-AM inequality, for any positive a , b , 2 a 2 + b 2 ≥ 2 a + b with equality if and only if a = b .
let x + 1 y + 1 z + 1 S T S 2 − T 2 = a , x + 2 = a 2 + 1 , a > 1 = b , y + 5 = b 2 + 4 , b > 1 = c , z + 1 0 = c 2 + 9 , c > 1 = a 2 + 1 + b 2 + 4 + c 2 + 9 = a + b + c = ( a 2 + 1 + b 2 + 4 + c 2 + 9 ) 2 − T 2 = 2 ( 2 a 2 + 1 2 + 2 b 2 + 4 + 2 c 2 + 9 ) 2 − T 2 ≥ 2 ( 2 a + 1 + 2 b + 2 + 2 c + 3 ) 2 − T 2 = 2 ( a + b + c + 6 ) 2 − ( a + b + c ) 2
with equality if and only if a = 1 , b = 2 , c = 3 , min ( S 2 − T 2 ) = 2 ( 1 + 2 + 3 + 6 ) 2 − ( 1 + 2 + 3 ) 2 = 3 6
note: there is a minor problem , to reach this minimum value, it must be a = 1 , x = 0 but x > 0 . so this minimum value is unreachable.
edit: ( x , y , z ) = ( 1 , 7 , 1 7 ) would make it to 3 6 . (thanks to Michael Mendrin)
Try ( x , y , z ) = ( 1 , 7 , 1 7 )
( 3 + 1 2 + 2 7 ) 2 − ( 2 + 8 + 1 8 ) 2 = 3 6
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good point. not sure where is the flaw in my reasoning though...
personally i think it's a huge problem, and it doesn't solve the problem. So abandon this way will be better :(
-Thanks for Cauchy-Schwarz S^2-T^2=(S+T)(S-T), clearly
Used Wolfram alpha to get the following four answers all giving 36 as the local minimum.
min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.100514, 3.40206, 8.90463)
min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.724565, 5.89826, 14.5211)
min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.567207, 5.26883, 13.1049)
min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.543924, 5.1757, 12.8953)
There might be are more such local minimum. I have printed four of them.
Answer=36
Use LaTeX; it is hard to read
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we let a = x + 1 , b = y + 1 , c = z + 1
then T = a + b + c ,and S = a 2 + 1 2 + b 2 + 2 2 + c 2 + 3 2
now consider this graph
Then A E = a 2 + 1 2 , E F = ( b 2 + 2 2 ) , F G = ( c 2 + 3 2 )
So T = A B + B C + C D = A D , a n d , S = A E + E F + F G
And then
S 2 − T 2 = ( A E + E F + F G ) 2 − A D 2 ≥ A G 2 − A D 2 = G D 2 = 3 6
when A,F,G in the same line the inequation holds.