一道中国高考原题，不服来做

we let $a=\sqrt{x+1},b=\sqrt{y+1},c=\sqrt{z+1}$

then $T=a+b+c$ ,and $S=\sqrt{a^2+1^2}+\sqrt{b^2+2^2}+\sqrt{c^2+3^2}$

now consider this graph

Then $AE=\sqrt{a^2+1^2},EF=\sqrt(b^2+2^2),FG=\sqrt(c^2+3^2)$

So $T = AB+BC+CD=AD, and,S = AE+EF+FG$

And then

$S^2-T^2 = (AE+EF+FG)^2-AD^2 \ge AG^2-AD^2=GD^2=36$

when A,F,G in the same line the inequation holds.

I think Cauchy-Schwarz is way more straightforward:

$\begin{aligned}S^2-T^2&=(S+T)(S-T)\\ &=\bigg[(\sqrt{x+2}-\sqrt{x+1})+(\sqrt{y+5}-\sqrt{y+1})+(\sqrt{z+10}-\sqrt{z+1})\bigg]\bigg[(\sqrt{x+2}+\sqrt{x+1})+(\sqrt{y+5}+\sqrt{y+1})+(\sqrt{z+10}+\sqrt{z+1})\bigg]\\ &=\bigg[(\sqrt{x+2}-\sqrt{x+1})+(\sqrt{y+5}-\sqrt{y+1})+(\sqrt{z+10}-\sqrt{z+1})\bigg]\bigg[\dfrac{1}{\sqrt{x+2}-\sqrt{x+1}}+\dfrac4{\sqrt{y+5}-\sqrt{y+1}}+\dfrac9{\sqrt{z+10}-\sqrt{z+1}}\bigg]\\&\geq (1+2+3)^2\\&=36\end{aligned}$

The equality holds when

$\begin{cases}\sqrt{x+2}-\sqrt{x+1}=1\\\sqrt{y+5}-\sqrt{y+1}=2\\\sqrt{z+10}-\sqrt{z+1}=3\end{cases}$ Which is accurately $x=y=z=0$ .

By cauchy inequality $\begin{aligned} S^{2}-T^{2} &=(S+T)(S-T)\\&=(\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}+\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1})(\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}-\sqrt{x+1}-\sqrt{y+1}-\sqrt{z+1})\\&=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+\sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1})(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}})\\&≧(1+2+3)^{2}\\&=\boxed{36} \end{aligned}$

Since $S^2-T^2=(S+T)(S-T)$ , the product rule gives us $\frac{d(S^2-T^2)}{dx} = 2S\frac{dS}{dx}-2T\frac{dT}{dx}$ .

At minimum this must be 0, so $\frac{S}{\sqrt{x+2}} = \frac{T}{\sqrt{x+1}}$ $\frac{S^2}{T^2} = \frac{x+2}{x+1}$ $\frac{S^2}{T^2} -1 = \frac{1}{x+1}$ Doing this for y and z as well we get $\frac{S^2}{T^2} -1 = \frac{1}{x+1} = \frac{4}{y+1} = \frac{9}{z+1}$ Introducing a constant k such that $\frac{1}{k}=\frac{S^2}{T^2} -1$ and using the original definition $T=\sqrt{1+x}+\sqrt{1+y}+\sqrt{1+z}$ , we can express T at minimum in terms of k: $T=\sqrt{k}+\sqrt{4k}+\sqrt{9k}=6\sqrt{k}$ $T^2=36k$

And using the definition for k, we find $S^2=(\frac{1}{k}+1)T^2=36+36k$ In the difference k does not appear, so we have our answer: $S^2-T^2=36$

Recall that by QM-AM inequality, for any positive $a,b$ , $\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2}$ with equality if and only if $a=b$ .

$\begin{aligned} \text{let } \sqrt{x+1} &= a , \sqrt{x+2} = \sqrt{a^2+1} , a>1 \\ \sqrt{y+1} &= b , \sqrt{y+5} = \sqrt{b^2+4} , b>1 \\ \sqrt{z+1} &= c , \sqrt{z+10} = \sqrt{c^2+9}, c>1 \\ \\ S &= \sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9} \\ T &= a+b+c \\ \\ S^2-T^2 &= (\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9})^2 - T^2 \\ &= 2 \left( \sqrt{\frac{a^2+1^2}{2}}+\sqrt{\frac{b^2+4}{2}}+\sqrt{\frac{c^2+9}{2}} \right)^2 - T^2 \\ &\geq 2 \left( \frac{a+1}{2}+\frac{b+2}{2}+\frac{c+3}{2} \right)^2 - T^2 \\ &= \frac{(a+b+c+6)^2}{2} - (a+b+c)^2 \end{aligned}$

with equality if and only if $a=1, b=2, c=3$ , $\min(S^2-T^2) = \frac{(1+2+3+6)^2}{2} - (1+2+3)^2 = 36$

note: there is a minor problem , to reach this minimum value, it must be $a=1, x=0$ but $x>0$ . so this minimum value is unreachable.

edit: $(x,y,z)=(1,7,17)$ would make it to $36$ . (thanks to Michael Mendrin)

-Thanks for Cauchy-Schwarz S^2-T^2=(S+T)(S-T), clearly

Used Wolfram alpha to get the following four answers all giving 36 as the local minimum.

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.100514, 3.40206, 8.90463)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.724565, 5.89826, 14.5211)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.567207, 5.26883, 13.1049)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.543924, 5.1757, 12.8953)

There might be are more such local minimum. I have printed four of them.

Answer=36