# 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.

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
3 years, 9 months ago

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

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

- 3 years, 9 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 $$\dfrac{1}{a^{2}} + \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}} \ge \dfrac{1}{ab} + \dfrac{1}{ac} + \dfrac{1}{bc}.$$

- 3 years, 9 months ago

Thats nice too.

- 3 years, 9 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,

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

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

- 3 years, 9 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 $$a,b,c>0$$

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

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

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

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

Updated! Thanks. :)

- 3 years, 9 months ago

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

- 3 years, 9 months ago