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Prove that $\forall a,b,c\ge 0$ then $$\sqrt{\frac{a^2+4bc}{b^2+c^2}}+\sqrt{\frac{b^2+4ac}{a^2+c^2}}+\sqrt{\frac{c^2+4ba}{b^2+a^2}}\ge 2+\sqrt{2}$$ I have some ideas but they didn't lead to simpler inequalities. I found the equality occurs for (a,b,c) = (1,1,0) Could you help me ? Thank you so much!

Arctic Char
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Frog WeII
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    Could you show some of your ideas? They may help..Note that the inequality occurs in general for $a=b,c=0$ – Albus Dumbledore Feb 26 '21 at 13:41
  • Standardized a+b+c = 2. Then Elimination (sorry because of my bad English) – Frog WeII Feb 26 '21 at 14:18
  • Mixing Variables – Frog WeII Feb 26 '21 at 14:28
  • Gradient is zero for $a=b=c$. The minimum is $3 \sqrt{\frac{5}{2}}\approx 4.74$ – Raffaele Feb 26 '21 at 14:47
  • @Raffaele When $a=b > 0, c=0$, the value is $2 + \sqrt{2}$. – River Li Feb 26 '21 at 23:47
  • @AlbusDumbledore did you found some ideas ? – Frog WeII Feb 27 '21 at 01:48
  • @FrogWeII No but i think we havfe to set some biginning inequality like nguyenhuyen_ag did here: https://math.stackexchange.com/questions/2057006/prove-that-sum-limits-cyc-sqrt-frac8ab8ac9bc2bcb2c-geq5?rq=1 – Albus Dumbledore Feb 27 '21 at 01:59
  • @RiverLi This means that Mathematica is wrong. Try FindMinimum[{(Sqrt[(b^2+4 a c)/(a^2+c^2)]+Sqrt[(4 a b+c^2)/(a^2+b^2)]+Sqrt[(a^2+4 b c)/(b^2+c^2)])&&a>=0&&b>=0&&c>=0},{a,b,c}] – Raffaele Feb 27 '21 at 12:11
  • @Raffaele You may add a constraint $a + b + c == 1$ in the code. FindMinimum[{(Sqrt[(b^2 + 4 a c)/(a^2 + c^2)] + Sqrt[(4 a b + c^2)/(a^2 + b^2)] + Sqrt[(a^2 + 4 b c)/(b^2 + c^2)]), a >= 0 && b >= 0 && c >= 0 && a + b + c == 1}, {a, b, c}] – River Li Feb 27 '21 at 12:41
  • @Raffaele However, if we add $a+b+c == 3$, does not work: FindMinimum[{(Sqrt[(b^2 + 4 a c)/(a^2 + c^2)] + Sqrt[(4 a b + c^2)/(a^2 + b^2)] + Sqrt[(a^2 + 4 b c)/(b^2 + c^2)]), a >= 0 && b >= 0 && c >= 0 && a + b + c == 3}, {a, b, c}] – River Li Feb 27 '21 at 12:46
  • @RiverLi I get your result with FindMinimum only if I set $c=0$ (or $a=0$ or $b=0$). – Raffaele Feb 27 '21 at 13:14
  • @Raffaele Can you copy my code? I used your code but it does not work. So I change your code a little bit. – River Li Feb 27 '21 at 13:40
  • @RiverLi with your code I get $${4.74342,{a\to 0.333333,b\to 0.333333,c\to 0.333333}}$$ – Raffaele Feb 27 '21 at 13:42
  • @Raffaele I get $${3.41422, {a -> 0.5, b -> 0.5, c -> 2.07714*10^-7}}$$ – River Li Feb 27 '21 at 13:44
  • Let us continue this discussion in chat. – Raffaele Feb 27 '21 at 13:47
  • Wolfram pro works perfect (but numerically) $≈3.41421$ minimize | function | sqrt((a^2 + 4 b c)/(b^2 + c^2)) + sqrt((b^2 + 4 a c)/(a^2 + c^2)) + sqrt((c^2 + 4 b a)/(b^2 + a^2)) domain | a>=0 ∧ b>=0 ∧ c>=0 – lone student Feb 27 '21 at 16:39
  • @lonestudent Yes, wolfram alpha also works but shows "4 local minima (no global minima found)": minimize(sqrt((a^2 + 4 b c)/(b^2 + c^2)) + sqrt((b^2 + 4 a c)/(a^2 + c^2)) + sqrt((c^2 + 4 b a)/(b^2 + a^2))) subject to a>=0,b>=0,c>=0 – River Li Feb 28 '21 at 05:52
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    @RiverLi see the super solution by arquady in aops https://artofproblemsolving.com/community/c6t243f6h260824_abcgt0 – Albus Dumbledore Mar 01 '21 at 10:31
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    @AlbusDumbledore Thanks. It is nice to eliminate the root sign. – River Li Mar 01 '21 at 10:47

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