# [SOLVED]Prove √a+√b+√c+√d≤10

#### anemone

##### MHB POTW Director
Staff member
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 $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.

#### lfdahl

##### Well-known member
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:

$\sqrt{a} +\sqrt{b} +\sqrt{c}+\sqrt{d} \leq 1+2+3+4 = 10.$

#### kaliprasad

##### Well-known member
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.

#### lfdahl

##### Well-known member
Thankyou for your comment. You´re right of course. I did miss something ...

#### DavidCampen

##### Member
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.

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#### anemone

##### MHB POTW Director
Staff member
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.