Since we have the following three inequalities for $a\geq b$: \[\sqrt{a^2+b^2}\leq a+b(\sqrt{2}-1)\] \[\sqrt[3]{a^3+b^3}\leq a+b(\sqrt[3]{2}-1)\] \[\sqrt[4]{a^4+b^4}\leq a+b(\sqrt[4]{2}-1)\] Therefore \[\sqrt{a^2+b^2}+\sqrt[3]{a^3+b^3}+\sqrt[4]{a^4+b^4} \le 3a+b(\sqrt{2}+\sqrt[3]{2}+\sqrt[4]{2}-3)<3a+b\]

Nice,gxggs! My solution is the same as your one.

thank you, very much!

Lemma: Let $a\geq b>0$ and $c_{n}={\frac{1}{\sqrt[n]{2}-1}}$ Then $\sqrt[n]{a^n+b^n}\leq a+\frac{b}{c_{n}}$ Proof: $(a+\frac{b}{c_{n}})^n=a^n+\sum_{i=1}^{n}\binom{n}{i}a^{n-i}\frac{b^i}{c^{i}_{n}}\geq a^n+\sum_{i=1}^{n}\binom{n}{i}\frac{b^n}{c^{i}_{n}}=a^n+b^n(\sum_{i=1}^{n}\binom{n}{i}\frac{1}{c^{i}_{n}})=a^n+b^n((1+\frac{1}{c_n})^n-1)=a^n+b^n$. Complete. We use lemma: \[\sqrt{a^2+b^2}\leq a+b(\sqrt{2}-1)\] \[\sqrt[3]{a^3+b^3}\leq a+b(\sqrt[3]{2}-1)\] \[\sqrt[4]{a^4+b^4}\leq a+b(\sqrt[4]{2}-1)\] Therefore \[\sqrt{a^2+b^2}+\sqrt[3]{a^3+b^3}+\sqrt[4]{a^4+b^4} \le 3a+b(\sqrt{2}+\sqrt[3]{2}+\sqrt[4]{2})-3b<3a+b(\sqrt{2.25}+\sqrt[3]{2.197}+\sqrt[4]{2.0736})-3b=3a+b(1.5+1.3+1.2)-3b=3a+b\]

Just a sidenote: Equality actually holds, when $a=b=0$

OO sorry. My solution For positive numbers

@ Sayan: Equality holds when $b=0$, obviously $a$ hasnt any restrictions anymore.

SCP wrote: @ Sayan: Equality holds when $b=0$, obviously $a$ hasnt any restrictions anymore. oh thanks. Didnt notice it

This problem was on Mathematical Reflections, i attached file

Attachments:
A nice lemma on inequalities.pdf (95kb)

Case 1: $b=0$ $3a=3a$. Case 2: $b>0$ let $x=\frac{a}{b}$ so $x\ge 1$ and we prove that $\sqrt{x^2+1}-\sqrt[3]{x^3+1}-\sqrt[4]{x^4+1}<3x+1$. Let $f(x)=3x+1-\sqrt{x^2+1}-\sqrt[3]{x^3+1}-\sqrt[4]{x^4+1}$. $f'(x)=3-\frac{x}{\sqrt{x^2+1}}-\frac{x^2}{\sqrt[3]{(x^3+1)^2}}-\frac{x^3}{\sqrt[4]{(x^4+1)^3}}>3-\frac{x}{\sqrt{x^2}}-\frac{x^2}{\sqrt[3]{(x^3)^2}}-\frac{x^3}{\sqrt[4]{(x^4)^3}}=0$ so $f(x)\ge f(1)=4-\sqrt{2}-\sqrt[3]{2}-\sqrt[4]{2}>0$

Case 1: $b=0$ $3a=3a$. Case 2: $b>0$ let $x=\frac{a}{b}$ so $x\ge 1$ and we prove that $\sqrt{x^2+1}+\sqrt[3]{x^3+1}+\sqrt[4]{x^4+1}<3x+1$. Let $f(x)=3x+1-\sqrt{x^2+1}-\sqrt[3]{x^3+1}-\sqrt[4]{x^4+1}$. $f'(x)=3-\frac{x}{\sqrt{x^2+1}}-\frac{x^2}{\sqrt[3]{(x^3+1)^2}}-\frac{x^3}{\sqrt[4]{(x^4+1)^3}}>3-\frac{x}{\sqrt{x^2}}-\frac{x^2}{\sqrt[3]{(x^3)^2}}-\frac{x^3}{\sqrt[4]{(x^4)^3}}=0$ so $f(x)\ge f(1)=4-\sqrt{2}-\sqrt[3]{2}-\sqrt[4]{2}>0$

Apmo problem

$\sqrt[2]{2}$ > $1$ and $3$ > $2\sqrt[2]{2}$ $\implies$ $(a + b)^2 \le a^2 + 2\sqrt[2]{2}ab + b^2 (3 - 2\sqrt[2]{2})$ $\implies$ $\sqrt[2]{a^2+b^2} \le a + b(\sqrt[2]{2} - 1)$ Similarly, We prove these: $\sqrt[3]{a^3+b^3} \le a + b(\sqrt[3]{2}-1)$ $ \sqrt[4]{a^4+b^4} \le a + b(\sqrt[4]{2}-1)$ We get $A=\sqrt{a^2+b^2}+\sqrt[3]{a^3+b^3}+\sqrt[4]{a^4+b^4}$ We add all $A \le 3a+b(\sqrt{2}+\sqrt[3]{2}+\sqrt[4]{2}-3)<3a+b$ $O.Y.SH.$

Let $a>b\ge 0$. Prove that $$ \frac{\sqrt{2a^2+b^2}- \sqrt[3]{a^3+2b^3}}{a+b} \leq \sqrt 2-1$$

Merlinlegion07's solution is wonderful