$m_1, m_2, m_3 $ are sides-lengths of a triangle such that $m_1\sqrt{m_1}+m_2\sqrt{m_2}=m_3\sqrt{m_3}$.
Prove that this triangle is an obtuse-angled triangle.
I don't have idea make run this example when I raised to square argument it makes a little bit complicated like this $m_1^3+2m_1\sqrt{m_1}m_2\sqrt{m_2}+m_2^3=m_3^3$
The hint.
Let $m_1=a$, $m_2=b$ and $m_3=c$.
Thus, $$c^2=\left(\sqrt{c^3}\right)^{\frac{4}{3}}=\left(\sqrt{a^3}+\sqrt{b^3}\right)^{\frac{4}{3}}>a^2+b^2.$$
The last inequality it's $$(\sqrt{a^3}+\sqrt{b^3})^4>(a^2+b^2)^3,$$ which is true because $$(\sqrt{a^3}+\sqrt{b^3})^4=a^6+4\sqrt{a^9b^3}+6a^3b^3+4\sqrt{a^3b^9}+b^6>$$ $$>a^6+3\sqrt{a^9b^3}+3\sqrt{a^3b^9}+b^6\geq a^6+3a^4b^2+3a^2b^4+b^6=(a^2+b^2)^3.$$
$$3\sqrt{a^9b^3}+3\sqrt{a^3b^9}\geq3a^4b^2+3ab^4$$ it's $$\sqrt{a^3b^3}(a^3+b^3)\geq a^2b^2(a^2+b^2)$$ or $$a^3+b^3\geq\sqrt{ab}(a^2+b^2)$$ or $$a^3-a^2\sqrt{ab}-b^2\sqrt{ab}+b^3\geq0$$ or $$a^2\sqrt{a}(\sqrt{a}-\sqrt{b})-b^2\sqrt{b}(\sqrt{a}-\sqrt{b})\geq0$$ or $$(\sqrt{a}-\sqrt{b})(\sqrt{a^5}-\sqrt{b^5})\geq0,$$ which is true because if $\sqrt{a}\geq\sqrt{b}$ so $\sqrt{a^5}\geq\sqrt{b^5}$ and if $\sqrt{a}\leq\sqrt{b}$ so $\sqrt{a^5}\leq\sqrt{b^5}$.