หาค่าคงที่ที่ดีที่สุด

[SOS_math]

ให้ $a,b,c>0$ จงหาค่าคงที่ $M$ (ในรูปของ $a,b,c$) ที่ดีที่สุดที่ทำให้
$$ax+by+cz\le M((x^2+y^2)^{1/2}+(y^2+z^2)^{1/2})$$
สำหรับทุก $x,y,z>0$

[Anton]

Problem. Let $a$, $b$, and $c$ be positive real numbers. Determine the best constant $M$ (in terms of $a$, $b$, and $c$) such that
$$ax+by+cz\leq M\bigg(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}\bigg)$$
for all positive real numbers $x$, $y$, and $z$.

If $a^2+b^2\leq c^2$, then $M=c$. This is because
$$ax+by+cz\leq \sqrt{a^2+b^2}\,\sqrt{x^2+y^2}+cz\leq c\,\sqrt{x^2+y^2}+cz$$
due to the Cauchy-Bunyakovsky-Schwarz Inequality. Thus,
$$ax+by+cz\leq c\,\sqrt{x^2+y^2}+c\,\sqrt{y^2+z^2}=c\,\left(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}\right)\,.$$
This shows that $M\leq c$. By fixing $z:=1$ and taking $x,y\to 0^+$, we note that $M\geq c$. Therefore, $M=c$ in this case.

Similarly, if $b^2+c^2\leq a^2$, then $M=a$. From now on, we assume that $a^2+b^2>c^2$ and $b^2+c^2>a^2$.

By the Cauchy-Bunyakovsky-Schwarz Inequality,
$$ax+\left(\frac{-a^2+b^2+c^2}{2b}\right)y\leq \sqrt{a^2+\left(\frac{-a^2+b^2+c^2}{2b}\right)^2}\,\sqrt{x^2+y^2}=\mu\,\sqrt{x^2+y^2}\,,$$
where
$$\mu:=\frac{\sqrt{a^4+b^4+c^4+2b^2c^2-2c^2a^2+2a^2b^2}}{2b}\,.$$
Similarly,
$$\left(\frac{a^2+b^2-c^2}{2b}\right)y+cz\leq \sqrt{\left(\frac{a^2+b^2-c^2}{2b}\right)^2+c^2}\,\sqrt{x^2+y^2}=\mu\,\sqrt{y^2+z^2}\,.$$
From
$$ax+by+cz=\left(ax+\left(\frac{-a^2+b^2+c^2}{2b}\right)^\vphantom{2^2}y\right)+\left(\left(\frac{a^2+b^2-c^2}{2b}\right)^\vphantom{2^2}y+cz\right)\,,$$
we conclude that
$$ax+by+cz\leq \mu\,\left(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}\right)\,.\tag{*}$$
This shows that $M\leq \mu$.

Now, we can take $$x=2ab(a^2+b^2-c^2)\,,$$ $$y=(-a^2+b^2+c^2)(a^2+b^2-c^2)\,,$$
and
$$z:=2bc(-a^2+b^2+c^2)\,.$$ Observe that (*) is an equality. Hence, $M\geq \mu$. Consequently, $M=\mu$. We conclude that
$$M=\begin{cases}a&\text{if }b^2+c^2\leq a^2\,,\\
c&\text{if }a^2+b^2\leq c^2\,,\\
\dfrac{\sqrt{a^4+b^4+c^4+2b^2c^2-2c^2a^2+2a^2b^2}}{2b}&\text{if }a^2+b^2>c^2\text{ and }b^2+c^2>a^2\,.\end{cases}$$