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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 $$\frac{1}{a^3(a+b^2)}+\frac{1}{b^3(b+c^2)}+\frac{1}{c^3(c+a^2)} \geqslant \frac{3}{2} $$

HN_NH
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2 Answers2

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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.

Michael Rozenberg
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  • Is it doable in hand :| :| – Rezwan Arefin Jan 17 '17 at 23:05
  • @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. – Michael Rozenberg Jan 17 '17 at 23:09
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    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? – Martin R Jan 20 '17 at 08:30
  • @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. – Michael Rozenberg Jan 20 '17 at 11:02
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    As a girl with honor, I kept my promise and reward the point. However, I do wish to see different solutions. – HN_NH Jan 31 '17 at 00:15
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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

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

Kiran
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