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

  • MHB
  • Thread starter anemone
  • Start date
In summary, the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$ can be proved using various methods such as mathematical induction, Cauchy-Schwarz inequality, or the AM-GM inequality, provided that $a,b,c,d$ are positive numbers. A specific value cannot be used to prove the inequality, as it must hold for all positive values. The inequality can be graphically represented by plotting the points $(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d})$ and is often used in fields such as economics and physics. It cannot be reversed to $\sqrt{a}+\sqrt{b}+\sqrt
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
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
  • #2
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:

$\sqrt{a} +\sqrt{b} +\sqrt{c}+\sqrt{d} \leq 1+2+3+4 = 10.$
 
  • #3
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.
 
  • #4
Thankyou for your comment. You´re right of course. I did miss something ... :(
 
  • #5
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:
  • #6
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.
 

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

What does the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$ mean?

The inequality means that the sum of the square roots of positive numbers a, b, c, and d is less than or equal to 10.

How do you prove the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$?

There are multiple ways to prove this inequality, but one approach is to use the Cauchy-Schwarz inequality. This states that for any positive numbers x and y, we have $(x+y)^2 \le 2(x^2+y^2)$. By applying this inequality to each pair of square roots, we can show that $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le \sqrt{2(a+b+c+d)}$. Then, using the fact that the square root function is concave, we can show that $\sqrt{2(a+b+c+d)} \le 10$, thus proving the original inequality.

Can the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$ be generalized to more than four terms?

Yes, the inequality can be generalized to any number of terms. In fact, it is a special case of the generalized Minkowski inequality, which states that for any positive numbers $x_1, x_2, ..., x_n$ and any positive real number p, we have $(x_1^p + x_2^p + ... + x_n^p)^{\frac{1}{p}} \le x_1 + x_2 + ... + x_n$. In the case of our inequality, we have p = 0.5 and n = 4.

What are some applications of the inequality $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$?

This inequality has various applications in mathematics and physics. For example, it can be used to prove the convergence of certain infinite series, as well as to establish the existence of solutions to certain differential equations. It also has applications in geometry, where it can be used to prove the existence of certain geometric constructions.

Are there any other similar inequalities involving square roots?

Yes, there are many other similar inequalities involving square roots. Some examples include the Cauchy-Schwarz inequality, the AM-GM inequality, and the Jensen's inequality. These inequalities have various applications in different areas of mathematics and can be used to prove many other important results.

Similar threads

MHB Prove Inequality: $\sqrt{ab}+\sqrt{cd}\le \sqrt{(a+d)(b+c)}$
Replies
1
Views
904
MHB Prove Triangle Sides Inequality $\sqrt{b+c-a} \dots \le 3$
Replies
1
Views
1K
MHB GRE.al.4 Find the domain of \sqrt{x^2-25}
Replies
3
Views
883
MHB Can You Prove this Trigonometric Inequality?
Replies
1
Views
915
MHB Inequality in Triangle: Prove $|x^2+y^2+z^2-2(xy+yz+xz)|\le \frac{1}{27}$
Replies
1
Views
1K
MHB Proving Inequality for Variables with Constraints
Replies
4
Views
2K
MHB Find a,b,c,d for Max $a+b-c+d$
Replies
1
Views
862
MHB Roots of equation and probability
Replies
1
Views
935
MHB Prove $\frac{1}{3}\le \frac{u}{v} \le \frac{1}{2}$ with Triangle Sides
Replies
4
Views
1K
MHB Inequality - find the largest K in (a+b+c+d)^2≥Kbc
Replies
7
Views
2K

Hot Threads

Recent Insights

Back
Top