Simo_the_Wolf wrote: Now for weighted CM-QM we have: ${( \frac{ a^{3}+b^{3}+2\sqrt{2}(\sqrt{ \frac{a^{2}+b^{2}}2 })^{3}}{1+1+2\sqrt{2}} )^{1/3}\geq ( \frac{ a^{2}+b^{2}+2\sqrt{2}(\sqrt{ \frac{a^{2}+b^{2}}2 })^{2}}{1+1+2\sqrt{2}} )^{1/2}= ( \frac{ a^{2}+b^{2}}2 } )^{1/2}$ Can anyone show me the complete form of weighted CM-QM?

If we do the good ol' $ a=y+z$ substitution, we have [wlog] $ (x+y)^{2}+(y+z)^{2}=(z+x)^{2}\implies x=\frac{(y)(y+z)}{z-y}$ [note $ z\ge y$! Plugging that in, we get: $ \frac{3y^{2}-2yz+z^{2}}{(4)(y+z)(z)}\ge k$ If we do $ z=y+t$ where $ t\ge 0$, we get: $ \frac{t^{2}+2y^{2}}{4(t+y)(t+2y)}\ge k$ If we set $ y=pt$, we get a function in $ p$ that can easily be minimized using calculus...the equality case follows as an isosceles right triangle.

let $ \angle A = 90$ we know that in any triangle we have: $ a^{3}+b^{3}+c^{3}= 2p(p^{2}-3r^{2}-6Rr)$ where $ r,R,p$ are the inradius and circumradius and semiperimeter respectively... now note that $ ABC$ is rightangled so $ a = 2R,r = p-a$ so the above inequality in our tirangle is equivalent to: $ a^{3}+b^{3}+c^{3}= 2p(p^{2}-3(p-a)^{2}-3a(p-a)) = 2p(-2p^{2}+3pa)$ also we have to find the biggest $ k$ for which the following inequality holds: $ 2p(-2p^{2}+3pa)\geq k(a+b+c)^{3}= 8kp^{3}$ $ \Rightarrow-2p^{2}+3pa\geq 4kp^{2}\rightarrow-2p+3a\geq 4kp\rightarrow 3a-a-b-c\geq 2k(a+b+c)$ $ \Rightarrow 2a-b-c\geq 2ka+2k(b+c)\rightarrow\boxed{2a(1-k)\geq (2k+a)(b+c) (I)}$ also we know that: $ a^{2}= b^{2}+c^{2}\rightarrow a =\sqrt{b^{2}+c^{2}}$ now from $ (I)$ we get that: $ \sqrt{b^{2}+c^{2}}\geq\frac{2k+1}{2-2k}(b+c)$ now let $ T =\frac{2k+1}{2-2k}$ its easy to show that $ k$ is maximum when $ T$ is maximum... now we know that: $ \sqrt{\frac{b^{2}+c^{2}}{2}}\geq\frac{b+c}{2}\rightarrow\sqrt{b^{2}+c^{2}}\geq\frac{\sqrt{2}}{2}(b+c)$ $ \Rightarrow T_{max}\geq\frac{\sqrt{2}}{2}$ but note that in the special case where $ \angle B =\angle C = 45$ we have that: $ b = c\rightarrow\sqrt{2b^{2}}\geq T(2b)\rightarrow T\leq\frac{\sqrt{2}}{2}$ so we get that: $ T_{max}=\frac{\sqrt{2}}{2}$ hence we get that: $ k_{max}=\frac{\sqrt{2}-1}{\sqrt{2}+2}$ which is the same result as Simo_the_Wolf...

BaBaK Ghalebi wrote: its easy to show that $ k$ is maximum when $ T$ is maximum... How ? explain your idea a little bit please

FOURRIER wrote: BaBaK Ghalebi wrote: its easy to show that $ k$ is maximum when $ T$ is maximum... How ? explain your idea a little bit please first of all note that $ k < 1$,now let $ T_1\geq T_2$,i.e. $ \frac {2k_1 + 1}{2 - 2k_1}\geq \frac {2k_2 + 1}{2 - 2k_2}$ then its sufficient to show that $ k_1\geq k_2$... now note that as we said before $ k_1,k_2 < 1$ so $ 2 - 2k_1,2 - 2K_2 < 0$ so $ T_1>T_2$ yields to: $ (2k_1 + 1)(2 - 2k_2)\geq (2k_2 + 1)(2 - 2k_1)$ which is equivalent to: $ k_1\geq k_2$ as wanted... now let $ T_{max} = \frac {2k' + 1}{2 - 2k'}$ then we know that $ T_{max}\geq T$ therefore $ k'\geq k$ thus $ k' = k_{max}$... so $ k$ is maximum whenever $ T$ gains it's maximum value...

Thank you Babak Ghalebi

One more thing: Why $ r = p - a$ if $ A = 90$? It is equivalent to $ r=\frac{b+c-a}{2}$ but...

FOURRIER wrote: One more thing: Why $ r = p - a$ if $ A = 90$? It is equivalent to $ r = \frac {b + c - a}{2}$ but... lemma.in triangle $ ABC$ let $ I$ be the incenter and let the circle touch the side $ AB$ at $ M$ then $ AM = p - a$. now let $ D,E$ be points on $ AB,AC$ s.t $ IM\perp AB,IN\perp AC$ so from $ \angle A = 90$ we get that $ AMIN$ is a square so $ r = AM$ now according to the lemma we get that $ AM = p - a$ so $ r = p - a$