# 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:

$a, b, c$ 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 :) Note by Trung Đặng Đoàn Đức
4 years, 5 months ago

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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 $a=b=c=\dfrac{1}{3}$.

- 4 years, 5 months ago

From $A.M \ge H.M$ 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, $a^2+b^2+c^2 \ge ab+bc+ca ~ \& \left( \dfrac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^{ 2 } } +\dfrac { 1 }{ c^{ 2 } } \right) \ge \left( \dfrac { 1 }{ ab } +\dfrac { 1 }{ bc } +\dfrac { 1 }{ ca } \right)$

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

- 4 years, 5 months ago

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

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

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

- 4 years, 4 months ago

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

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

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

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

- 4 years, 5 months ago

Thats nice too.

- 4 years, 5 months ago

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

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 $a,b,c>0$

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

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

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

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Updated! Thanks. :)

- 4 years, 5 months ago