$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ \sum_{cyc}\frac{a^2}{ca^2 + 2c^2} \ge 1$$   Here's what I did. My stupidity has reached a spiritual level.  We have that $ab + bc + ca = 3abc \implies \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 3$.  $$\sum_{cyc}\dfrac{a^2}{ca^2 + 2c^2} = \sum_{cyc}\dfrac{1}{c}\left(1 - \dfrac{2c}{a^2 + 2c}\right) \ge \sum_{cyc}\dfrac{1}{c}\left(1 - \dfrac{2c}{2c + 2a - 1}\right)$$  $$ = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) - 2\sum_{cyc}\dfrac{1}{2c + 2a - 1} \ge 3 - \dfrac{2}{9}\sum_{cyc}\left(\dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{c + a - 1}\right)$$  $$ = 3 - \dfrac{4}{9}\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) - \dfrac{2}{9}\sum_{cyc}\dfrac{1}{c + a - 1} \ge 3 - \dfrac{4}{3} - \dfrac{1}{18}\sum_{cyc}\left(\dfrac{1}{c - \frac{1}{2}} + \dfrac{1}{a - \frac{1}{2}}\right)$$  $$ = \dfrac{5}{3} - \dfrac{2}{9}\left(\dfrac{1}{2a - 1} + \dfrac{1}{2b - 1} + \dfrac{1}{2c -1}\right)$$  I am done with my life.
let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$   show that   $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$   This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you
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Some story about these unusual little blue electric critters that only certain people could see, and these kids took it upon themselves to go about destroying them since they fed off energy or something like that.  Not sure whether they fed off people or off electricity... or both maybe?  I think the dominant method of destroying them was an electric baton that was taken from law enforcement.  I wanna call them buzz batons, but that's probably thanks to Artemis Fowl. I know that if we solve the Maxwell equation, we will end up with the phase velocity of light being related to the permeability and the permittivity of the material. But this is not what I'm interested in - I want to go deeper than that. We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with $c$ in the spacing. The rare atoms will disturb the light in some way. So I am interested in how the atoms affect the light. Photon absorption-emission theory Some textbooks that I read explain it in a way kind of like this: In a material the photons are absorbed by atom and then re-emitted a short time later, then they travel a short distance to the next atom and get absorbed&amp;emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the apparent speed of light in that material. So the light appears slower because it has a smaller “drift speed”. Interference theory But recently I realize an alternative explanation: Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that results in delayed light (advanced in phase), this can easily be shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second, which gives the impression of a lower phase velocity. However the group velocity is changing in a complicated way. 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So the question is What really cause the phase velocity of light to be decreased?  &quot;drift velocity&quot; of photons (they aren't the same photons, they are re-emitted all the time) phase difference between absorbed and emitted light something else  And also, I still don't really understand detailed explanation of the absorption-emission process for small light's wavelength (for large lambda compare to the atoms spacing, the photons will be absorbed by the phonons). The dispersion relation that we know is continuous and also some material is non-dispersive, therefore the absorption process must occur in all frequency for a certain range. So definitely it doesn't involve the atomic transition, otherwise it will be quantized. My guess is that the relevant absorption process gets smooth out by the dipole moment. What makes the spectrum continuous? EDIT: link for dispersion relation: Right identity and Right inverse implies a group