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Prove or disprove that the inequality $$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$ is valid if $x,y,z$ are positive numbers and $$xyz=1.$$

My solution is: Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\dfrac{c}{a}.$$ So we have $$\dfrac{1}{\sqrt{1+\dfrac{a}{b}}}+\dfrac{1}{\sqrt{1+\dfrac{b}{c}}}+\dfrac{1}{\sqrt{1+\dfrac{c}{a}}}=\dfrac{1}{\sqrt{\dfrac{a+b}{b}}}+\dfrac{1}{\sqrt{\dfrac{b+c}{c}}}+\dfrac{1}{\sqrt{\dfrac{c+a}{a}}}=\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}=\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}}.$$ Then we can use this: $$\dfrac{1}{\sqrt{xy}}\geq\dfrac{2}{x+y}.$$ So we have $$\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}} \geq \dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c}.$$ Then we can use this: $$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \geq 3 \sqrt[3]{\dfrac{abc}{xyz}}.$$ So we have $$\dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c} \geq 3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}}.$$ The question is what should I do next? Is it already obvious that $$3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}} \geq 1?$$ Any hint would help a lot! Thanks!

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  • $\begingroup$ Note that your final claim isn't true, eg for $ a = 100, b = 1, c = 1$. $\endgroup$
    – Calvin Lin
    Commented Dec 31, 2022 at 16:44

2 Answers 2

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Let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.

Thus, $z=\frac{c}{a}$ and $$\sum_{cyc}\frac{1}{\sqrt{1+x}}=\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a+b}}\geq\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$

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  • $\begingroup$ Why is this inequality valid? $\endgroup$ Commented Dec 22, 2022 at 17:39
  • $\begingroup$ @Alice Malinova because $\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}\geq\sqrt{a+b}.$ $\endgroup$ Commented Dec 22, 2022 at 17:41
  • $\begingroup$ okay, but so far I don't see why there is $$\sqrt{c}$$ in the first fraction? Isn't just $$\dfrac{1}{\sqrt(a+b) \geq \dfrac{1}{\sqrt{a}+\sqrt{b}}?$$ $\endgroup$ Commented Dec 22, 2022 at 17:47
  • $\begingroup$ @Alice Malinova $\frac{1}{\sqrt{1+x}}=\frac{1}{\sqrt{1+\frac{a}{b}}}=\frac{\sqrt{b}}{\sqrt{a+b}}\geq\frac{\sqrt{b}}{\sqrt{a+b+2\sqrt{ab}}+\sqrt{c}}=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$ $\endgroup$ Commented Dec 22, 2022 at 18:19
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    $\begingroup$ now I see, thank you very much for help! $\endgroup$ Commented Dec 23, 2022 at 14:51
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This is almost the same as the other answer and just slightly different: $\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}\ge\sqrt{\dfrac{b}{a+b+c}}+\sqrt{\dfrac{c}{a+b+c}}+\sqrt{\dfrac{a}{a+b+c}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+b+c}}> 1.$ Done.

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  • $\begingroup$ so we can just add \sqrt{c} in the first denominator (etc) to make our right fraction less than left? $\endgroup$ Commented Dec 23, 2022 at 1:08
  • $\begingroup$ No, you add $c$ only. Not $\sqrt{c}$. $\endgroup$
    – Wang YeFei
    Commented Dec 23, 2022 at 1:49
  • $\begingroup$ ok, now I see, thanks a lot! $\endgroup$ Commented Dec 23, 2022 at 14:51

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