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shorter proof of generalized mediant inequality?

A geometric proof without words : But with words, it's better. Let $V_k=\pmatrix{a_k\\b_k}$ and $\theta_k \in I$ its polar angle where $I=(0,\tfrac{\pi}{2})$ The idea is that it suffices to "...
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shorter proof of generalized mediant inequality?

Assume $$\frac{a_1}{b_1}\le\frac{a_2}{b_2}\le\dots\le\frac{a_n}{b_n}$$ and define $$N_k:=a_1+a_2+\dots+a_k\quad\text{and}\quad D_k:=b_1+b_2+\dots+b_k.$$ Using the ordinary mediant inequality, ...
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Accepted

• 105

Young's inequality for scalar multiplied with absolute value of function

I believe the following holds for $x,y \in \mathbb{R}$ because $|x|^2 = x^2$ and for $y$ as awell. $$x\cdot y \leq |x|\cdot |y| \leq \frac{1}{2\cdot \epsilon}\cdot x^2 + \frac{\epsilon}{2}\cdot y^2$$...
• 36
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Prove that $a^2+b^2+c^2 \geq ab(a+b+\sqrt{ab})+cb(c+b+\sqrt{cb})+ ac(a+c+\sqrt{ac} )$

Multiply LHS by $a+b+c=1$. You get after expansion: $$\sum a^3 + \sum(a^2b+ab^2)\ge \sum (a^2b+ab^2) +\sum a^{3/2}b^{3/2},$$ which is true due to Muirhead or just $x^2+y^2+z^2\ge xy+yz+zx.$
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Equality in Hardy's inequality via Hölder's

I would like to suggest this approach. Let's $C_C$ be the set of continuous compact supported functions and $C_C^+$ the subset of non negative functions of $C_C$. For all $f\in C_C^+$, it's easy (...
• 316
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Minimizing $2a^2 + b^2 + c^2$ given $4a + 3b + c = 7$

Hint, Let $f(a,b,c)=2a^2+b^2+c^2$ and $g(a,b,c)=4a+3b+c-7$ , Now consider $\nabla f=\lambda \nabla g$ with $g(a,b,c)=0$.
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