Thanks to Olena Bormashenko http://www.kalva.demon.co.uk/apmo/asoln/asol034.html

I don't know why I can't open the link

Just prove if$0<x\leq y$,then $\sqrt[n]{x^n+y^n}\leq(\sqrt[n]{2}-1)x+y$ Then it's easy.

$ x=\dfrac{-a+b+c}{2}$ ; $ y=\dfrac{a-b+c}{2}$ ; $ z=\dfrac{a+b-c}{2}$ By Minkowsky: $ \sqrt[n]{(y+z)^n+(x+z)^n}+\sqrt[n]{(z+x)^n+(y+x)^n}+\sqrt[n]{(x+y)^n+(z+y)^n}\le (\sqrt[n]{y^n+x^n}+\sqrt[n]{z^n+z^n})+(\sqrt[n]{z^n+y^n}+\sqrt[n]{x^n+x^n})+(\sqrt[n]{x^n+z^n}+\sqrt[n]{y^n+y^n})$ $ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}\le \sqrt[n]{x^n+y^n}+\sqrt[n]{y^n+z^n}+\sqrt[n]{z^n+x^n}+\sqrt[n]{2}(x+y+z)$ $ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}\le \sqrt[n]{x^n+y^n}+\sqrt[n]{y^n+z^n}+\sqrt[n]{z^n+x^n}+\dfrac{\sqrt[n]{2}}{2}$ And because: $ x^n+y^n<(x+y)^n$ $ \Rightarrow$ $ \sqrt[n]{x^n+y^n}<x+y$ $ y^n+z^n<(y+z)^n$ $ \Rightarrow$ $ \sqrt[n]{y^n+z^n}<y+z$ $ z^n+x^n<(z+x)^n$ $ \Rightarrow$ $ \sqrt[n]{z^n+x^n}<z+x$ Then: $ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<(x+y)+(y+z)+(z+x)+\dfrac{\sqrt[n]{2}}{2}$ $ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<2(x+y+z)+\dfrac{\sqrt[n]{2}}{2}$ $ \boxed{\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\dfrac{\sqrt[n]{2}}{2}}$

$ \mbox{ My solutions of the Jin's inequality, in attachment: }$

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MARIAN DINCA proof.pdf (71kb)

shobber wrote: Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that \[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \] We may assume that $a \le b \le c < \frac 12$. Then \begin{align*} \sqrt[n]{b^n+a^n} &< b + \frac1n \frac{a^n}{b^{n-1}} < b + \frac1n a \\ \sqrt[n]{c^n+a^n} &< c + \frac1n \frac{a^n}{c^{n-1}} < c + \frac1n a \\ \sqrt[n]{b^n+c^n} &< \sqrt[n]{\left( \frac12 \right)^n + \left( \frac12 \right)^n} = \frac{\sqrt[n]{2}}{2}. \end{align*} Summing and using the fact $n \ge 2$ yields the conclusion.

shobber wrote: Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that \[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \] n=2,Romania 1983

Let $a$, $b$ and $c$ be sides-lengths of triangle such that $a+b+c=1$. Prove that:$$\sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{b}^{2}}+{{c}^{2}}}+\sqrt{{{c}^{2}}+{{a}^{2}}}+\frac{abc}{ab+bc+ca}\le 1+\frac{1}{\sqrt{2}}$$

shobber wrote: Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that \[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \] WLOG. $c\le b\le a<\frac{1}{2}$. Claim. $b^n+a^n<\left(b+\frac{c}{2}\right)^n$. Proof. $b^n+c^n<b^n+b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}+\cdots+\left(\frac{c}{2}\right)^n$ $c^n<b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}+\cdots+\left(\frac{c}{2}\right)^n$ See $n\ge 2$ such that, $b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}=b^{n-1}\times\frac{c}{2}\times n>b^{n-1}\times c$ $c^n<b^{n-1}\times c$ QED. Next, we get : $\sqrt[n]{a^n+b^n}\le\sqrt[n]{a^n+a^n}=a\sqrt[n]{2}<\frac{\sqrt[n]{2}}{2}$ $\sqrt[n]{b^n+a^n}<b+\frac{c}{2}$ $\sqrt[n]{a^n+c^n}<a+\frac{c}{2}$ Thus, we get $\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<a+b+c+\frac{\sqrt[n]{2}}{2}=1+\frac{\sqrt[n]{2}}{2}$. QED

Nice problem , But hard Lemma : if $x \geq 1$ then $\sqrt[n]{x^n+1} \leq x+\sqrt[n]{2}-1$ . Prove : We can see $f^{'}(x) = \frac{x^{n-1}}{(x^n+1)^{\frac{n-1}{n}}}-1 <0$ So $f$ is descending function . Then we can see : $$f(x) \leq f(1)=\sqrt[n]{2}-1$$ Now if we let $m \geq j >0$ , $x=\frac{m}{j}$ . Then $\sqrt[n]{m^n+j^n} \leq m+(\sqrt[n]{2}-1)j$ . ( ) Now let $a \geq b \geq c$ . Now we can see : $$\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n} \leq \sqrt[n]{2}(a+b+c) = \sqrt[n]{2} < 1+\frac{ \sqrt[n]{2}}{2}$$

strong_boy wrote: Nice problem , But hard Lemma : if $x \geq 1$ then $\sqrt[n]{x^n+1} \leq x+\sqrt[n]{2}-1$ . Prove : We can see $f^{'}(x) = \frac{x^{n-1}}{(x^n+1)^{\frac{n-1}{n}}}-1 <0$ So $f$ is descending function . Then we can see : $$f(x) \leq f(1)=\sqrt[n]{2}-1$$ Now if we let $m \geq j >0$ , $x=\frac{m}{j}$ . Then $\sqrt[n]{m^n+j^n} \leq m+(\sqrt[n]{2}-1)j$ . ( ) Now let $a \geq b \geq c$ . Now we can see : $$\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n} \leq \sqrt[n]{2}(a+b+c) = \sqrt[n]{2} < 1+\frac{ \sqrt[n]{2}}{2}$$ nice problem-beautiful solution!

Let $a\ge b\ge c$. $\ \ \ 1=a+b+c>2a \Rightarrow \ \ a<\frac{1}{2}$ $\sqrt[n]{{{a}^{n}}+{{b}^{n}}}\le \sqrt[n]{2a^n}=a\cdot \sqrt[n]{2}<\frac{\sqrt[n]{2}}{2}$ For $0<x\le1\Rightarrow \ \ \sqrt[n]{1+x^n}\le \sqrt{1+x^n}\le \sqrt{1+x^2}<1+\frac{x}{2}$ $\Rightarrow \ \ \sqrt[n]{1+(\frac{c}{a})^n}<1+\frac{c}{2a}, \ \ \sqrt[n]{1+(\frac{c}{b})^n}<1+\frac{c}{2b}$ $\Rightarrow \ \ \sqrt[n]{a^n+c^n}<a+\frac{c}{2}, \ \ \sqrt[n]{b^n+c^n}<b+\frac{c}{2}$ $\Rightarrow \ \ \sqrt[n]{a^n+c^n}+\sqrt[n]{b^n+c^n}<(a+\frac{c}{2})+(b+\frac{c}{2})=1$ $\Rightarrow \ \ \sqrt[n]{a^n+c^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{a^n+b^n}<1+\frac{\sqrt[n]{2}}{2}$

shobber wrote: Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that \[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \]

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