2001 IMO SL #A6

Last one. I promise. 

 Prove that for all positive real numbers $a,b,c$,\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.  \]

Solution 1:

Note that this inequality is homogeneous. WLOG, assume $a+b+c=1$.

Now, we can proceed by using Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$. This gives us:

\[\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}\]

Now, we finish off with AM-GM:

\[\displaystyle 1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc\]

And this proves the inequality.

Solution 2:

By Cauchy-Schwarz, we have

\[ (\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2 (\sum_{cyc}{a(a^2+8bc)}).\]

\[ \geq(a+b+c)^3\]

And we know that 

\[ (a+b+c)^3\geq\sum_{cyc}{a(a^2+8bc)},\]

so we obtain

\[ (\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}})^2  \geq\frac{(a+b+c)^3}{(\sum{a(a^2+8bc)})}\]

\[ \geq1,\]

as desired. $\square$

Solution 3:

From Holder's, $(\sum(\frac{a}{\sqrt{a^2+8bc}}))^2(a(a^2+8bc)+b(b^2+8ca)+c(c^2+8ab))\ge (a+b+c)^3$.

Now, we have $(\sum(\frac{a}{\sqrt{a^2+8bc}}))^2 \ge \frac{(a+b+c)^3}{a(a^2+8bc)+b(b^2+8ca)+c(c^2+8ab)}$.

We have our desired $\sum \frac{a}{\sqrt{a^2+8bc}}\ge 1\iff \frac{(a+b+c)^3}{a(a^2+8bc)+b(b^2+8ca)+c(c^2+8ab)}\ge 1$ from above.

Therefore, we now have to prove $\frac{(a+b+c)^3}{a(a^2+8bc)+b(b^2+8ca)+c(c^2+8ab)}\ge 1$. We desire $a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\ge a^3+b^3+c^3+24abc$.

Therefore, we now want $3a^2b+3ab^2+3a^2+3ac^2+3b^2c+3bc^2\ge 18abc$ which is true by AM-GM, so we are done. $\square$

Solution 4: 

By Holder's,$$\left(\sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}} \right)^2 \left(\sum_{cyc} a(a^2 + 8bc) \right) \ge (a+b+c)^3$$Since $a,b,c$ are positive, we need to show that$$(a+b+c)^3\geq \sum_{cyc} \frac{a}{\sqrt{a^2 + 8bc}}$$Expanding and simplifying gives us the following:

$$a^3+b^3+c^3+3(\sum_{cyc}a^2b)+6abc\geq a^3+b^3+c^3+24abc$$$$\sum_{cyc}a^2b\geq 6ab$$This is true because it is Muirhead's on $(2, 1, 0)\succ(1,1,1)$, and we are done. $\square$

Solution 5:

Firstly, $a^{2}+8bc=(a^{2}+2bc)+6bc \leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc$ (where $S=a^{2}+b^{2}+c^{2}$) and its cyclic variations.

Next note that $(a,b,c)$ and $\left( \frac{1}{\sqrt{S+6bc}}, \frac{1}{\sqrt{S+6ca}}, \frac{1}{\sqrt{S+6ab}} \right)$ are similarly oriented sequences. Thus

$$\sum_{cyc} \frac{a}{\sqrt{a^{2}+8bc}} \ge \sum_{cyc} \frac{a}{\sqrt{S+6bc}}$$$$\geq ^{chev} \frac{1}{3}(a+b+c)\left( \frac{1}{\sqrt{S+6bc}}+\frac{1}{\sqrt{S+6ca}}+\frac{1}{\sqrt{S+6ab}} \right)$$$$\geq^{AH} \frac{1}{3}(a+b+c) \left( \frac{9}{\sqrt{S+6bc}+\sqrt{S+6ca}+\sqrt{S+6ab}} \right)$$$$\geq^{QA} (a+b+c) \sqrt{\frac{3}{(S+6bc)+(S+6ca)+(S+6ab)}}$$$$=(a+b+c)\sqrt{\frac{3}{3(a+b+c)^{2}}}=1$$

Notation: $AG$: AM-GM inequality, $AH$: AM-HM inequality, $chev$: Chebyshev inequality, $QA$: QM-AM inequality aka RMS inequality


At last, I'm somewhat satisfied.  

I would do another problem but I promised you that this is the last one so... that's it I guess?


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