A problem 2.

[Hojoo Lee]

Let $a,b,c \geqslant 0$ and no two of which are zero. Prove that
\[\sum_{cyc}\sqrt{1+\frac{48a}{b+c}} \geqslant 15\]

[nooonuii]

ข้อนี้ต้องลบด้วยรึเปล่าครับ

[tatari/nightmare]

ไม่ต้องครับ ข้อนี้ไม่ใช่ข้อสอบ
source of Problem : Crux???? solution by Li ZHou

[Anonymous314]

เฉลยมันใช้ เกินความรู้ิอะครับ

[Hojoo Lee]

Solution by Li Zhou
Without loss of generality, we may assume that $a\geqslant b\geqslant c\geqslant 0$ and
that $a+b+c=1$. Let $f(x)=\sqrt{1+\frac{48x}{1-x}}$ for $0\leqslant x< 1$. Then
\[f'(x)=\frac{24}{\sqrt{(1-x)^3(1+47x)}}>0\]
and \[f''(x)=\frac{48(47x-11)}{\sqrt{(1-x)^5(1+47x)^3}}.\]
The tangen line to the graph of $f(x)$ at the point $(\frac{1}{3},5)$ has equation
$T(x)=\frac{54+7}{5}$. Setting $f(x)=T(x)$, we obtain $12(3x-1)^2(27x-2)=0$,from
which we see that the graph of the functions $f$ and $T$ intersect again at
$x=\frac{2}{27}$. Define
\[g(x)=y = \cases{T(x) if \frac{2}{27}\leqslant x<\frac{1}{3},\cr f(x) if \frac{1}{3}\leqslant x<1.} \]
Clearly, the function $g$ is convex and $g(x)\leqslant f(x)$ for $\frac{2}{27}\leqslant x<1$.
if $b\leqslant \frac{2}{27}$, then $a=1-b-c\geqslant \frac{23}{27}$, and therefore,
\[f(a)\geqslant f(\frac{23}{27})=\sqrt{277}>15\],
which implies that the original inequality holds. Hence, we can further assume that
$b>\frac{2}{27}$.
if $c>\frac{2}{27}$, then, applying Jensen's Inequality, we obtain
\[f(a)+f(b)+f(c)\geqslant g(a)+g(b)+g(c)\]
\[\geqslant 3g\left(\,\frac{1}{3}(a+b+c)\right) =3g\left(\,\frac{1}{3}\right) =15,\]
with equality if and only if $a=b=c=\frac{1}{3}$.
if $\frac{1}{17}<c\leqslant \frac{2}{27}$, then $f(c)>f(\frac{1}{27})=2$ and, applying
Jensen's Inequality, we obtain
\[f(a)+f(b)\geqslant g(a)+g(b)\geqslant 2g\left(\,\frac{1}{2}(a+b)\right) \geqslant 2g\left(\,\frac{25}{54}\right) >13.\]
Thus, $f(a)+f(b)+f(c)>15$, and the original inequality holds agains.
Finally, consider $c\leqslant \frac{1}{17}$. Then, applying Jensen's Inequality, we have
\[f(a)+f(b)+f(c)\geqslant g(a)+g(b)+f(c)\geqslant 2g\left(\,\frac{1}{2}(a+b)\right) +f(c)\]
\[=2f\left(\,\frac{1}{2}(a+b)\right) +f(c)=2f\left(\,\frac{1}{2}(1-c)\right) +f(c).\]
Define
\[h(x)=2f\left(\,\frac{1-x}{2}\right) +f(x)=2\sqrt{\frac{49-47x}{1+x}}+\sqrt{\frac{1+47x}{1-x}}\]
for $0\leqslant x\leqslant \frac{1}{17}$. Then $h(0)=15$ and
\[h'(x)=24\left(\,\frac{1}{\sqrt{(1-x)^3(1+47x)}}-\frac{4}{\sqrt{(1+x)^3(49-47x)}}\right).\]
Now, $(1+x)^3(49-47x)-16(1-x)^3(1+47x)=(3x-1)k(x)$, where $k(x)=235x^3-699x^2+505x-33$.
It is easy to verify that $k\left(\,\frac{1}{17}\right) <0$, $k(1)=0$, $k\left(\,\frac{3}{2}\right) <0$,
and $k(2)>0$. Hence, $k(x)<0$ for $0\leqslant x\leqslant \frac{1}{17}$. Thus, $h'(x)>0$,
and therefore, $h(x)\geqslant h(0)=15$ for $0\leqslant x\leqslant \frac{1}{17}$.
This completes the proof. In summary, equality holds if and only if $a=b=c$ or two of
$a,b,c$ are equal while the third is 0.

[tatari/blackman]

เก่งจังเลยครับคุณ Hojoo Lee

[God Phoenix]

นี่เป็นเฉลยใน crux รึเปล่าครับ???

[nooonuii]

ใช่ึครับ ยากสุดๆไปเลย

[The jumpers]

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