MHB Prove $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$ w/ $a,b,c,d>0$

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion focuses on proving the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$ under the conditions that $a, b, c, d > 0$, $a \le 1$, $a + b \le 5$, $a + b + c \le 14$, and $a + b + c + d \le 30$. Participants explore various mathematical approaches and inequalities, such as the Cauchy-Schwarz inequality, to validate the claim. The constraints on the variables are critical in deriving the upper limit of the sum of square roots. The proof ultimately hinges on the relationships between the variables and their upper bounds. The conclusion affirms that the inequality holds true given the specified conditions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that if $a,\,b,\,c,\,d>0$ and $a\le 1,\,a+b\le 5,\,a+b+c\le 14,\,a+b+c+d\le 30$, then $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.
 
Mathematics news on Phys.org
Did I miss something, or is this problem really quite easy to solve? Thankyou for any comment!

Given the conditions:

$a \leq 1 \;\;\wedge \;\;a+b \leq 5\;\; \wedge \;\;a+b+c\leq 14\;\; \wedge \;\; a+b+c+d \leq 30$

which by successive subtractions implies:

$a \leq 1 \;\;\wedge \;\;b \leq 4\;\; \wedge \;\;c\leq 9\;\; \wedge \;\; d \leq 16$

or:

$\sqrt{a} \leq 1 \;\;\wedge \;\;\sqrt{b} \leq 2\;\; \wedge \;\;\sqrt{c}\leq 3\;\; \wedge \;\; \sqrt{d} \leq 4$

Adding the four inequalities yields:

$\sqrt{a} +\sqrt{b} +\sqrt{c}+\sqrt{d} \leq 1+2+3+4 = 10.$
 
lfdahl said:
Did I miss something, or is this problem really quite easy to solve? Thankyou for any comment!

Given the conditions:

$a \leq 1 \;\;\wedge \;\;a+b \leq 5\;\; \wedge \;\;a+b+c\leq 14\;\; \wedge \;\; a+b+c+d \leq 30$

which by successive subtractions implies:

$a \leq 1 \;\;\wedge \;\;b \leq 4\;\; \wedge \;\;c\leq 9\;\; \wedge \;\; d \leq 16$

or:

$\sqrt{a} \leq 1 \;\;\wedge \;\;\sqrt{b} \leq 2\;\; \wedge \;\;\sqrt{c}\leq 3\;\; \wedge \;\; \sqrt{d} \leq 4$

Adding the four inequalities yields:

$\sqrt{a} +\sqrt{b} +\sqrt{c}+\sqrt{d} \leq 1+2+3+4 = 10.$
a = .2 b = 4.6 does not satisfy your consideration.
 
Thankyou for your comment. You´re right of course. I did miss something ... :(
 
anemone said:
Prove that if $a,\,b,\,c,\,d>0$ and $a\le 1,\,a+b\le 5,\,a+b+c\le 14,\,a+b+c+d\le 30$, then $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.

Given the constraints on the 4 partitions of the value 30 it is true that the maximum value of the sum of the square roots of these partitions is 10. The problem is proving it.

It can be shown that the sum of the square roots of 2 partitions of a value is greater than or equal to the square root of the unpartitioned value and that the value of the sum of the square roots of the 2 partitions is maximized when the partitions are equal. The equation is:
$\sqrt{a}+\sqrt{b} = \sqrt{a+b+2\sqrt{ab}}$
let the value being partitioned = 1 then a+b=1 and b=a-1 and the equation becomes:
$\sqrt{a}+\sqrt{b} = \sqrt{1 + 2\sqrt{a(1-a)}}$
and it can be seen that as $a$ ranges from 0 to 1, the maximum value of the sum of the square roots of the two partitions is at $a$ = 1/2. We should be able to use induction to show that whatever the values of the 2 partitions, any additional partitions of the original two partitions can only increase the value of the sum of the square roots of all the partitions. Applying this to the constraining inequalities given for the four partitions we find that
a=1, b=4, c=9 and d=16 give the maximal value for the sum of the square roots of these four partitions within the given constraints and that this sum is 10 which demonstrates the assertion.
 
Last edited:
The function $f: (0,\,+\infty)\rightarrow (0,\,+\infty)$ defined by $f(x)=\sqrt{x}$ is concave, and therefore for any positive real numbers $k_1,\,k_2,\,\cdots, \,k_n$ such that $k_1+k_2+\cdots+k_n=1$, we have

$k_1f(x_1)+k_2f(x_2)+\cdots+k_nf(x_n)\le f(k_1x_1+k_2x_2+\cdots+k_nx_n)$

Now, take $n=4$ and $k_1=\dfrac{1}{10},\,k_2=\dfrac{2}{10},\,k_3=\dfrac{3}{10},\,k_4=\dfrac{4}{10}$. It follows that

$\dfrac{1}{10}\sqrt{a}+\dfrac{2}{10}\sqrt{\dfrac{b}{4}}+\dfrac{3}{10}\sqrt{\dfrac{c}{9}}+\dfrac{4}{10}\sqrt{\dfrac{d}{16}}\le \sqrt{\dfrac{a}{10}+\dfrac{b}{20}+\dfrac{c}{30}+\dfrac{d}{40}}$

or

$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10\sqrt{\dfrac{12a+6b+4c+3d}{120}}$

But

$12a+6b+4c+3d=3(a+b+c+d)+(a+b+c)+2(a+b)+6a\le 3(30)+14+2(5)+6(1)=120$

and the claim is then proved.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
3
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
6
Views
1K
Replies
1
Views
2K
Back
Top