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Problem (ARML 1987): If $a,b,c$ are positive and $a+b+c=6$, show that $$\left(a + \frac{1}{b} \right)^2+\left(b + \frac{1}{c} \right)^2 + \left(c + \frac{1}{a} \right)^2\geq \frac{75}{4}.$$

Workings

We want to prove

$$a^2+b^2+c^2+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{75}{4}$$ given $a+b+c=6$.

By AM-GM and with manipulation, we get the following (equivalent) inequalities: \begin{align} \frac{a+b+c}{3}=2&\geq\sqrt[3]{abc} \\ 4&\geq\sqrt[3]{a^2b^2c^2} \\ \frac{1}{4}&\geq \frac{1}{\sqrt[3]{a^2b^2c^2}}. \end{align}

Additionally, \begin{align} a^2+b^2+c^2&\geq3\sqrt[3]{a^2b^2c^2} \\ 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)&\geq 6 \\ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}&\geq 3\frac{1}{\sqrt[3]{a^2b^2c^2}}. \end{align}

Summing them together, we get

\begin{align} \left(a + \frac{1}{b} \right)^2+\left(b + \frac{1}{c} \right)^2 + \left(c + \frac{1}{a} \right)^2 &\geq 3\sqrt[3]{a^2b^2c^2}+3\frac{1}{\sqrt[3]{a^2b^2c^2}}+6 \quad(1) \\ \frac{75}{4}&\geq 3\sqrt[3]{a^2b^2c^2}+3\frac{1}{\sqrt[3]{a^2b^2c^2}}+6\quad(2) \end{align}

Question

Can I make the jump to say $$\left(a + \frac{1}{b} \right)^2+\left(b + \frac{1}{c} \right)^2 + \left(c + \frac{1}{a} \right)^2\geq \frac{75}{4}$$

because (1) is always satisfied for all values of RHS, and the maximum of RHS is $\frac{75}{4}$ (from eq. 2)?

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  • $\begingroup$ No, you cannot do that. $\endgroup$ Dec 31, 2023 at 9:48
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    $\begingroup$ For example, $7\geq 3$ and $10 \geq 3$, but it isnt true that $7 \geq 10$. $\endgroup$ Dec 31, 2023 at 9:50
  • $\begingroup$ But what if $a\geq b$ for all $b$ and $10 \geq b$? Doesn't this imply $a\geq 10$ is always true for any $b$? $\endgroup$ Dec 31, 2023 at 9:57
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    $\begingroup$ $0 \geq -x^2$ and $1 \geq -x^2$ does not imply $0 \geq 1$ $\endgroup$ Dec 31, 2023 at 9:59
  • $\begingroup$ An example closer to the OP's: $-x^2+1 \geq -x^2$ for all real $x$; the maximum of $-x^2$ is $0$; but it doesn't mean that $-x^2+1 \geq 0$. $\endgroup$
    – Litho
    Dec 31, 2023 at 19:45

2 Answers 2

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Here is my solution:

Dividing both sides by $3$ and square rooting, we yield

$$\sqrt{\frac{(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2}{3}} \geq \frac{5}{2}$$

But we also have by QM-AM that

$$\sqrt{\frac{(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2}{3}} \geq \frac{a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}}{3} = \frac{a+b+c+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}}{3}$$

We can rewrite this as $$2+\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}$$

Thus, we need to prove that $$2+\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{5}{2}$$

or

$$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{1}{2}$$

But from Titu's lemma (special case of Cauchy Schwarz inequality) we have that

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{(1+1+1)^2}{a+b+c} = \frac{9}{6}=\frac{3}{2}$$

Thus, we have

$$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{\frac{3}{2}}{3}=\frac{1}{2}$$

and we are done.

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  • $\begingroup$ Slightly more direct to use AM-HM to show that $ 1/a+1/b+1/c \geq 9/(a+b+c) = 3/2$. $\endgroup$
    – Calvin Lin
    Dec 31, 2023 at 20:48
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As indicated in the comments, your original approach is incorrect.

It would have worked if the condition was $abc = 8$, in which case it made sense to combine them as the product.

Since the condition is $ a+b+c = 6$, we slightly modify your AM-GM to using QM-AM-GM-HM to show that:

  1. $(a^2 +b^2 + c^2) \geq \frac{(a+b+c)^2}{3} = \frac{6^2}{3} = 12$
  2. $ 2( \frac{a}{b} + \frac{b}{c} + \frac{c}{a}) \geq 6 $
  3. $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{3^3}{(a+b+c)^2} = \frac{3}{4}$

Summing them up, we are done.

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  • $\begingroup$ What does the third inequality stem from? $\endgroup$ Dec 31, 2023 at 22:09
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    $\begingroup$ @RobertMurray QM-HM applied on $(1/a, 1/b, 1/c)$, which gives us $ \sqrt{ (1/a^2 + 1/b^2 + 1/c^2 ) / 3 } \geq 3/ ( a + b+ c ) $. $\endgroup$
    – Calvin Lin
    Jan 1 at 2:22

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