This could be the noobiest note ever posted on Brilliant :)

I'm about to have the final-semester test and here is one of the practice questions:

a,b,ca, b, c are positive real numbers such that a+b+c=1a+b+c=1.

Prove that (a+1a)2+(b+1b)2+(c+1c)2>33 { (a+\dfrac { 1 }{ a } ) }^{ 2 }+{ (b+\dfrac { 1 }{ b } ) }^{ 2 }+{ (c+\dfrac { 1 }{ c } ) }^{ 2 }>33.

I know that the minimum of the above equation is 3009\dfrac {300}{9} where a=b=c=13a=b=c=\dfrac {1}{3} but I don't know how to prove it.

Please help me, test is going on very soon :)

Note: Done the test, there is a very nice question in the test too. I'll post it some time later :)

Note by Trung Đặng Đoàn Đức
4 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Here is another method;

\because For positive real numbers A.MG.MA.M \ge G.M and equality holds if and only if all numbers are equal.

\therefore For minimum value of (a+1a)2+(b+1b)2+(c+1c)2(a+1a)2=(b+1b)2=(c+1c)2 { (a+\dfrac { 1 }{ a } ) }^{ 2 }+{ (b+\dfrac { 1 }{ b } ) }^{ 2 }+{ (c+\dfrac { 1 }{ c } ) }^{ 2 } \\ {(a+\dfrac { 1 }{ a } ) }^{ 2 } ={ (b+\dfrac { 1 }{ b } ) }^{ 2 } = {(c+\dfrac { 1 }{ c } ) }^{ 2 }

So, a2+1a2+2=b2+1b2+2    a2b2+1a21a2=0     (ab)(a+b)(11a2b2)=0 ab=0 or a+b=0 or ab=1 or ab=1a^2+\dfrac{1}{a^2}+2=b^2+\dfrac{1}{b^2}+2 \\ \implies a^2-b^2 +\dfrac{1}{a^2} - \dfrac{1}{a^2}=0 \\ \implies ~ (a-b)(a+b)(1-\dfrac{1}{a^2b^2})=0 \\ \therefore ~ a-b=0~\text{or}~ a+b=0 ~ \text{or} ~ ab=1 ~\text{or}~ ab=-1

Now a+ba+b can never be 0 because both aa & bb are positive so abab would not be 1-1 for the same reason. Only thing left is to eliminate our fourth root i.e. ab=1ab = 1.

We know that a+b+c=1a+b+c=1. So a+b<1a+b<1.

Now (a+b)2=a2+b2+2ab      (a+b)2>2ab     1>2abab<12(a+b)^2 = a^2 + b^2 + 2ab ~ \\ \implies ~ (a+b)^2>2ab ~ \implies 1>2ab \\ \therefore ab< \dfrac{1}{2}.

So only root left is a=ba=b. Similarly b=cb=c & c=ac=a.

 a=b=c\therefore ~ a=b=c

Hence to minimise (a+1a)2+(b+1b)2+(c+1c)2 { (a+\dfrac { 1 }{ a } ) }^{ 2 }+{ (b+\dfrac { 1 }{ b } ) }^{ 2 }+{ (c+\dfrac { 1 }{ c } ) }^{ 2 } the condition is a=b=c=13a=b=c=\dfrac{1}{3}.

Purushottam Abhisheikh - 4 years, 5 months ago

Log in to reply

From A.MH.MA.M \ge H.M we get (1a+1b+1c)9a+b+c(1a+1b+1c)9\left( \dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right) \ge \dfrac{9}{a+b+c} \\ \therefore \left( \dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right) \ge 9

(a+b+c)2+(1a+1b+1c)282    a2+b2+c2+2(ab+bc+ac)+(1a2+1b2+1c2)+2(1ab+1bc+1ca)82\therefore (a+b+c)^2 + \left( \dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right)^2 \ge 82 \\ \implies a^2+b^2+c^2 + 2(ab+bc+ac) + \left( \dfrac { 1 }{ a^2 } +\dfrac { 1 }{ b^2 } +\dfrac { 1 }{ c^2 } \right) + 2\left( \dfrac { 1 }{ ab } +\dfrac { 1 }{ bc } +\dfrac { 1 }{ ca } \right) \ge 82

By Cauchy Schwarz inequality, a2+b2+c2ab+bc+ca &(1a2+1b2+1c2)(1ab+1bc+1ca)a^2+b^2+c^2 \ge ab+bc+ca ~ \& \left( \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge \left( \dfrac { 1 }{ ab } +\dfrac { 1 }{ bc } +\dfrac { 1 }{ ca } \right)

3(a2+b2+c2+1a2+1b2+1c2)82    (a2+b2+c2+1a2+1b2+1c2)823    (a2+b2+c2+1a2+1b2+1c2+6)823+6    (a+1a)2+(b+1b)2+(c+1c)2823+6>33\therefore 3\left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge 82 \\ \implies \left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge \dfrac{82}{3} \\ \implies \left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } +6 \right) \ge \dfrac{82}{3} + 6 \\ \implies \left( a+\dfrac{1}{a} \right)^2+\left( b+\dfrac{1}{b} \right)^2+\left( c+\dfrac{1}{c} \right)^2 \ge \dfrac{82}{3} + 6 > 33.

Purushottam Abhisheikh - 4 years, 5 months ago

Log in to reply

Using Jensen's inequality trivializes it. First, we note that the function f(x)=(x+1x)2f(x)=\left(x+\dfrac{1}{x}\right)^2 is convex  x>0\forall~x\gt 0 using second derivative test. Now, we use Jensen on f(x)f(x) with x=a,b,cx=a,b,c to get,

f(a)+f(b)+f(c)3×f(a+b+c3)=3×f(13)=3×(13+3)2=1003>993=33f(a)+f(b)+f(c)\geq 3\times f\left(\frac{a+b+c}{3}\right)=3\times f\left(\frac{1}{3}\right)=3\times \left(\frac{1}{3}+3\right)^2=\frac{100}{3}\gt \frac{99}{3}=33

f(a)+f(b)+f(c)>33\therefore\quad f(a)+f(b)+f(c)\gt 33

Prasun Biswas - 4 years, 4 months ago

Log in to reply

Nice proof. Rather than the Cauchy-Schwarz inequality I used the Chebyshev Sum Inequality. Without loss of generality, we can choose a,b,ca,b,c in descending order, in which case the inequality tells us that

3(a2+b2+c2)(a+b+c)(a+b+c)=a2+b2+c2+2(ab+ac+bc)3(a^{2} + b^{2} + c^{2}) \ge (a + b + c)(a + b + c) = a^{2} + b^{2} + c^{2} + 2(ab + ac + bc)

2(a2+b2+c2)2(ab+ac+bc)a2+b2+c2ab+ac+bc.\Longrightarrow 2(a^{2} + b^{2} + c^{2}) \ge 2(ab + ac + bc) \Longrightarrow a^{2} + b^{2} + c^{2} \ge ab + ac + bc.

Similarly for the inequality 1a2+1b2+1c21ab+1ac+1bc.\dfrac{1}{a^{2}} + \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}} \ge \dfrac{1}{ab} + \dfrac{1}{ac} + \dfrac{1}{bc}.

Brian Charlesworth - 4 years, 5 months ago

Log in to reply

Thats nice too.

Purushottam Abhisheikh - 4 years, 5 months ago

Log in to reply

(a+1a)2+(b+1b)2+(c+1c)2=(a2+b2+c2)+(1a2+1b2+1c2)+6(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2=(a^2+b^2+c^2)+(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})+6

Prove that 3(x2+y2+z2)(x+y+z)23(x^2+y^2+z^2)\geq (x+y+z)^2 with all x,y,zx,y,z

a2+b2+c2(a+b+c)2/3=1/3\Rightarrow a^2+b^2+c^2\geq (a+b+c)^2/3=1/3

(a+b+c)(1a+1b+1c)9(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9 with all a,b,c>0a,b,c>0

1a+1b+1c9(a+b+c=1)\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9(a+b+c=1)

1a2+1b2+1c2(1a+1b+1c)2/3>=92/3=27\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2/3>=9^2/3=27

(1a2+1b2+1c2)+(a2+b2+c2)+627+1/3+6>33\Rightarrow (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})+(a^2+b^2+c^2)+6\geq 27+1/3+6>33

(a+1/a)2+(b+1/b)2+(c+1/c)2>33\Rightarrow (a+1/a)^2+(b+1/b)^2+(c+1/c)^2>33

Duy Vu - 4 years, 5 months ago

Log in to reply

Hello, I have edited the LaTeX\LaTeX in your comment. Can you please check it out for accuracy?

Pranjal Jain - 4 years, 5 months ago

Log in to reply

It should be (x2+y2+z2)(x^2+y^2+z^2) in the second line from top.

Purushottam Abhisheikh - 4 years, 5 months ago

Log in to reply

@Purushottam Abhisheikh Updated! Thanks. :)

Pranjal Jain - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...