Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

Áp dụng BĐT AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)

Ta thấy 
72
=
2
3
.
3
2
72=2 
3
 .3 
2
  nên a, b có dạng 
{
�
=
2
�
3
�
�
=
2
�
.
3
�
{ 
a=2 
x
 3 
y
 
b=2 
z
 .3 
t
 
​
  với 
�
,
�
,
�
,
�
∈
N
x,y,z,t∈N và 
�
�
�
{
�
,
�
}
=
3
;
�
�
�
{
�
,
�
}
=
2
max{x,z}=3;max{y,t}=2. 

 Theo đề bài, ta có 
2
�
.
3
�
+
2
�
.
3
�
=
42
2 
x
 .3 
y
 +2 
z
 .3 
t
 =42

 
⇔
2
�
−
1
.
3
�
−
1
+
2
�
−
1
3
�
−
1
=
7
⇔2 
x−1
 .3 
y−1
 +2 
z−1
 3 
t−1
 =7   (*), do đó 
�
,
�
,
�
,
�
≥
1
x,y,z,t≥1

 TH1: 
�
≥
�
,
�
≤
�
x≥z,y≤t. Khi đó 
�
=
3
,
�
=
2
x=3,t=2. (*) thành:

 
4.
3
�
−
1
+
3.
2
�
−
1
=
7
4.3 
y−1
 +3.2 
z−1
 =7 
⇔
�
=
�
=
1
⇔y=z=1

 Vậy 
{
�
=
24
�
=
18
{ 
a=24
b=18
​
  (nhận)

 TH2: KMTQ thì giả sử 
�
≥
�
,
�
≥
�
x≥z,y≥t. Khi đó 
�
=
3
,
�
=
2
x=3,z=2. (*) thành 

 
4.
3
�
−
1
+
2.
3
�
−
1
=
7
4.3 
y−1
 +2.3 
t−1
 =7, điều này là vô lí.

 Vậy 
(
�
,
�
)
=
(
24
,
18
)
(a,b)=(24,18) hay 
(
18
,
24
)
(18,24) là cặp số duy nhất thỏa yêu cầu bài toán.

\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

\(P=\dfrac{6}{2xy+2yz+2zx}+\dfrac{2}{x^2+y^2+z^2}\ge\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=8+4\sqrt{3}\)