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Here's my question and my approach in latex since I can't type in mathjax.

https://mathb.in/76931

Show that for all positive real numbers $a, b, $ and $c:$ $$a \sqrt{bc} + b \sqrt{ca} + c \sqrt{ab} \leq \frac{1}{3}(a + b + c)^2$$ My approach: We know that $(a-\sqrt{bc})^2 + (b-\sqrt{ca})^2 + (c-\sqrt{ab})^2 \geq 0$. Then $a^2 + b^2 + c^2 + ab + ca + bc \geq 2a\sqrt{bc} + 2b\sqrt{ca} + 2c\sqrt{ab}$. Pretty close but I have no clue how to continue the proof.

Math Attack
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Adrien
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    FYI To type in mathjaxj, just change your brackets to dollar signs. I edited your writeup into the statement $\quad$ So all that's left is to show that $ ab + bc + ca \geq a \sqrt{bc} + b \sqrt{ca} + c\sqrt{ab}$. Are you familiar with AM-GM, if so think how about each term on the RHS could be viewed as a product of terms on the LHS, so that we can use AM-GM. – Calvin Lin Nov 16 '23 at 15:56
  • For your first question, yes I'm familiar with the AM-GM inequality, yet I have no clue how can I use it in that case using it in that case – Adrien Nov 16 '23 at 16:06
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    Humor me for a moment, can you write out an AM-GM proof for $a^2 +b^2 + c^2 \geq ab + bc + ca$? – Calvin Lin Nov 16 '23 at 16:07
  • I can show it using rearrangement like this: $(a-b)^2+(b-c)^2+(c-a)^2\geq 0$ meaning that: $2(a^2+b^2+c^2)\geq 2(ab+ca+bc)$ Hence $a^2+b^2+c^2\geq ab+ca+bc$ – Adrien Nov 16 '23 at 16:12
  • Well, If you did it via AM-GM like I asked, it would be realizing that $a^2 + b^2 \geq 2ab, b^2 + c^2 \geq 2bc, c^2 + a^2 \geq 2ac$, and if we sum those up we get $a^2+b^2+c^2 \geq ab+bc+ca$. $\quad$ Given that, can you use this approach of "each term on the RHS could be viewed as a product of terms on the LHS" to solve the "left to show that"? – Calvin Lin Nov 16 '23 at 16:18
  • Still no idea, ig I'm pretty dumb at this point – Adrien Nov 16 '23 at 16:30
  • $b\sqrt{ac} = \sqrt{ b^2 ac } = \sqrt{ (ba)(bc) } $, so we apply $ ab + bc \geq 2 \sqrt{ab^2c } = b\sqrt{ac} $. – Calvin Lin Nov 16 '23 at 16:34
  • Oh, now I see, tysm – Adrien Nov 16 '23 at 16:44
  • can you write that up as a solution? – Calvin Lin Nov 16 '23 at 16:55

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Since $\text{AM}\geq \text{GM}$

$a^2+b^2\geq 2ab$, $b^2+c^2\geq 2bc$, $c^2+a^2 \geq2ac$

that is

$a^2+b^2+c^2\geq ab+bc+ac$

and

$ab+ac\geq 2a\sqrt{bc}$, $ab+bc\geq 2b\sqrt{ac}$, $ac+bc\geq 2c\sqrt{ab}$

that is

$ab+bc+ac\geq a\sqrt{bc}+ b\sqrt{ac}+c\sqrt{ab} $

then

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\geq3(ab+bc+ac)\geq3(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab})$

Lion Heart
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