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

\(a, b, c\) are positive real numbers such that \(a+b+c=1\).

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

I know that the minimum of the above equation is \(\dfrac {300}{9}\) where \(a=b=c=\dfrac {1}{3}\) 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 :)

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

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

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

\(\therefore 3\left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge 82 \\ \implies \left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge \dfrac{82}{3} \\ \implies \left(a^2+b^2+c^2+ \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } +6 \right) \ge \dfrac{82}{3} + 6 \\ \implies \left( a+\dfrac{1}{a} \right)^2+\left( b+\dfrac{1}{b} \right)^2+\left( c+\dfrac{1}{c} \right)^2 \ge \dfrac{82}{3} + 6 > 33\). – Purushottam Abhisheikh · 2 years, 3 months ago

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Chebyshev Sum Inequality. Without loss of generality, we can choose \(a,b,c\) in descending order, in which case the inequality tells us that

Nice proof. Rather than the Cauchy-Schwarz inequality I used the\(3(a^{2} + b^{2} + c^{2}) \ge (a + b + c)(a + b + c) = a^{2} + b^{2} + c^{2} + 2(ab + ac + bc)\)

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

Similarly for the inequality \(\dfrac{1}{a^{2}} + \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}} \ge \dfrac{1}{ab} + \dfrac{1}{ac} + \dfrac{1}{bc}.\) – Brian Charlesworth · 2 years, 3 months ago

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– Purushottam Abhisheikh · 2 years, 3 months ago

Thats nice too.Log in to reply

\[f(a)+f(b)+f(c)\geq 3\times f\left(\frac{a+b+c}{3}\right)=3\times f\left(\frac{1}{3}\right)=3\times \left(\frac{1}{3}+3\right)^2=\frac{100}{3}\gt \frac{99}{3}=33\]

\[\therefore\quad f(a)+f(b)+f(c)\gt 33\] – Prasun Biswas · 2 years, 3 months ago

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

Prove that \(3(x^2+y^2+z^2)\geq (x+y+z)^2\) with all \(x,y,z\)

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

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

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

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

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

\(\Rightarrow (a+1/a)^2+(b+1/b)^2+(c+1/c)^2>33 \) – Duy Vu · 2 years, 3 months ago

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– Pranjal Jain · 2 years, 3 months ago

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

– Purushottam Abhisheikh · 2 years, 3 months ago

It should be \((x^2+y^2+z^2)\) in the second line from top.Log in to reply

– Pranjal Jain · 2 years, 3 months ago

Updated! Thanks. :)Log in to reply

Here is another method;

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

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

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

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

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

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

So only root left is \(a=b\). Similarly \(b=c\) & \(c=a\).

\(\therefore ~ a=b=c\)

Hence to minimise \( { (a+\dfrac { 1 }{ a } ) }^{ 2 }+{ (b+\dfrac { 1 }{ b } ) }^{ 2 }+{ (c+\dfrac { 1 }{ c } ) }^{ 2 } \) the condition is \(a=b=c=\dfrac{1}{3}\). – Purushottam Abhisheikh · 2 years, 3 months ago

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Second, I think I know where you're wrong. If we plug those values into LHS, \(a+ \dfrac {1}{a} = \dfrac {10}{3}\), not \(\dfrac {4}{3}\). (Encountered this mistake too :p) – Trung Đặng Đoàn Đức · 2 years, 3 months ago

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– Brian Charlesworth · 2 years, 3 months ago

Yes, I realized I made a silly mistake so I deleted my comment. The inequality is correct; I checked using Lagrange multipliers, (i.e., calculus), but I'm still trying to figure out which inequality theorem might be useful here so as to find a non-calculus solution. Sorry, but I'll have to get back to this problem later when I have more time.Log in to reply