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$a,b,c >0$ and $abc=1$, prove $$\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$$
1. I tried rearrangement and AM-GM but fail.
2. I think the power of $\frac14$ is tough. I can prove the easier inequality $$\frac{1}{a^3(a+b^2)}+\frac{1}{b^3(b+c^2)}+\frac{1}{c^3(c+a^2)} \geqslant \frac{3}{2} $$

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  • $\begingroup$ What about trying to raise the inequality to the power of 4 and going from there $\endgroup$ Dec 27, 2016 at 7:07

2 Answers 2

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Dedicated to Dr. Sonnhard Graubner.

Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ be positive numbers.

Hence, we need to prove that $$\sum_{cyc}\frac{1}{\sqrt[4]{\frac{y^3}{x^3}\left(\frac{y}{x}+\frac{z^2}{y^2}\right)}}\geq\frac{3}{\sqrt[4]2}$$ or $$\sum_{cyc}\frac{x}{\sqrt[4]{y(y^3+z^2x)}}\geq\frac{3}{\sqrt[4]2}.$$ By Holder $$\left(\sum_{cyc}\frac{x}{\sqrt[4]{y(y^3+z^2x)}}\right)^4\sum_{cyc}xy(y^3+z^2x)(x+z)^5\geq\left(\sum_{cyc}(x^2+xz)\right)^5.$$ Thus, it remains to prove that $$2\left(\sum_{cyuc}(x^2+xy)\right)^5\geq81\sum_{cyc}xy(y^3+z^2x)(x+z)^5.$$ Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$. Hence, $$2\left(\sum_{cyuc}(x^2+xy)\right)^5-81\sum_{cyc}xy(y^3+z^2x)(x+z)^5=$$ $$=3024(u^2-uv+v^2)x^8+144(83u^3-54u^2v-27uv^2+83v^3)x^7+$$ $$+48(401u^4-25u^3v-519u^2v^2+164uv^3+401v^4)x^6+$$ $$+2(7597u^5+7895u^4v-15650u^3v^2-9980u^2v^3+11864uv^4+7597v^5)x^5+$$ $$+(7015u^6+15482u^5v-6085u^4v^2-35170u^3v^3+64u^2v^4+18398uv^5+7015v^6)x^4+$$ $$(2078u^7+6595u^6v+4803u^5v^2-14295u^4v^3-7815u^3v^4+9663u^2v^5+6919uv^6+2078v^7)x^3+$$ $$+(380u^8+1597u^7v+2492u^6v^2-1234u^5v^3-3500u^4v^4+2006u^3v^5+2978u^2v^6+1597uv^7+380v^8)x^2+$$ $$+(40u^9+200u^8v+479u^7v^2+311u^6v^3-301u^5v^4+509u^4v^5+635u^3v^6+479u^2v^7+200uv^8+40v^9)x+$$ $$+2u^{10}+10u^9v+30u^8v^2+60u^7v^3+9u^6v^4+102u^5v^5+90u^4v^6+60u^3v^7+30u^2v^8+10uv^9+2u^{10}\geq0.$$ Done!

Thank you HN_NH for your down vote.

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  • $\begingroup$ Is it doable in hand :| :| $\endgroup$ Jan 17, 2017 at 23:05
  • $\begingroup$ @Rezwan Arefin I think the answer is "yes", but I used WA for the last step. By the way, we can prove the last inequality by hand after full expanding. $\endgroup$ Jan 17, 2017 at 23:09
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    $\begingroup$ Why is the last expression $\ge 0$? Is it "obvious" that all expressions in parentheses (like $7597u^5+7895u^4v−15650u^3v^2−9980u^2v^3+11864uv^4+7597v^5$) are non-negative? $\endgroup$
    – Martin R
    Jan 20, 2017 at 8:30
  • $\begingroup$ @Martin R For example, $7597x^5+7895x^4-15650x^3-9980x^2+11864x+7597=(7597x^5-15650x^3+11864x)+(7895x^4-9980x^2+7597)>0$ by AM-GM. $\endgroup$ Jan 20, 2017 at 11:02
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    $\begingroup$ As a girl with honor, I kept my promise and reward the point. However, I do wish to see different solutions. $\endgroup$
    – HN_NH
    Jan 31, 2017 at 0:15
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From AM–GM inequality

$\dfrac{\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\dfrac{1}{\sqrt[4]{b^3(b+c^2)}}+\dfrac{1}{\sqrt[4]{c^3(c+a^2)}}}{3} \geq \sqrt[3]{\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}\times\frac{1}{\sqrt[4]{c^3(c+a^2)}}}$

$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{a^3(a+b^2)b^3(b+c^2)c^3(c+a^2)}}}$

$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(abc)^3(a+b^2)(b+c^2)(c+a^2)}}}$

$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}} \cdots(1)$

From AM–GM inequality

$\dfrac{(a+b^2)+(b+c^2)+(c+a^2)}{3}\ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)}$

$\dfrac{(a+b+c)+(a^2+b^2+c^2)}{3}\ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)} > \cdots(2)$

From AM–GM inequality

$\dfrac{a+b+c}{3}\ge \sqrt[3]{abc}$

$\dfrac{a+b+c}{3}\ge \sqrt[3]{1}$

$a+b+c \ge 3 \cdots(3)$

From AM–GM inequality

$\dfrac{a^2+b^2+c^2}{3}\ge \sqrt[3]{a^2b^2c^2}$

$\dfrac{a^2+b^2+c^2}{3}\ge \sqrt[3]{(abc)^2}$

$\dfrac{a^2+b^2+c^2}{3}\ge 1$

$a^2+b^2+c^2 \ge 3 \cdots(4)$

Consider the equations $(3)$ and $(4)$. From these, we have

$(a+b+c)+(a^2+b^2+c^2) \ge 3+3$

$(a+b+c)+(a^2+b^2+c^2) \ge 6$

$\dfrac{(a+b+c)+(a^2+b^2+c^2)}{3} \ge 2 \cdots(5)$

Consider the equations $(2)$ and $(5)$. From these, we have

$2 \ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)}$

$8 \ge (a+b^2)(b+c^2)(c+a^2)$

$\sqrt[4]{8} \ge \sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}$

$\dfrac{1}{\sqrt[4]{8}} \le \dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}$

$\left(\dfrac{1}{\sqrt[4]{8}}\right)^{1/3} \le \sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$

$3\left(\dfrac{1}{\sqrt[4]{8}}\right)^{1/3} \le 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$

$\dfrac{3}{\left(\sqrt[4]{8}\right)^{1/3}} \le 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$

$\dfrac{3}{\sqrt[4]{2}} \le 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}} ~~\cdots(6)$

Consider the equations $(1)$ and $(6)$. From these, we have

$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\dfrac{1}{\sqrt[4]{b^3(b+c^2)}}+\dfrac{1}{\sqrt[4]{c^3(c+a^2)}} \ge \dfrac{3}{\sqrt[4]{2}}$

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    $\begingroup$ Although I find it exceedingly verbose (I mean, look at the derivation of 3,4 and 5...), this answer is in my opinion both simple and really the one that explains why the inequality holds. $\endgroup$
    – Anonymous
    Jan 30, 2017 at 22:32
  • $\begingroup$ The proof is fault, btw $\endgroup$
    – HN_NH
    Jan 31, 2017 at 0:14
  • $\begingroup$ Why the proof is fault? $\endgroup$
    – Kiran
    Jan 31, 2017 at 4:25

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