由于在之前提到过恒等式证明不等式的厉害之处，因此这里便再介绍一道可以改用换元解法解决的不等式题。
a > 0 , b > 0 , c > 0 , a b c = 1 ，求证： a + 3 ( a + 1 ) 2 + b + 3 ( b + 1 ) 2 + c + 3 ( c + 1 ) 2 ≥ 3. a>0, b>0, c>0, abc=1\text{，求证：}\frac{a+3}{\left( a+1 \right) ^2}+\frac{b+3}{\left( b+1 \right) ^2}+\frac{c+3}{\left( c+1 \right) ^2}\ge 3. a>0,b>0,c>0,abc=1，求证：(a+1)2a+3+(b+1)2b+3+(c+1)2c+3≥3.
在证明之前，还有一个重要的恒等式需要说明：
若 a > 0 , b > 0 , c > 0 , a b c = 0 , 则令 x = a − 1 a + 1 , y = b − 1 b + 1 , z = c − 1 c + 1 ，有： x y z + x + y + z = 0. \text{若}a>0, b>0, c>0, abc=0, \text{则令}x=\frac{a-1}{a+1}, y=\frac{b-1}{b+1}, z=\frac{c-1}{c+1}\text{，有：}\\xyz+x+y+z=0. 若a>0,b>0,c>0,abc=0,则令x=a+1a−1,y=b+1b−1,z=c+1c−1，有：xyz+x+y+z=0.
证明倒也不难，即直接代入通分：
证明： x y z + x + y + z = a − 1 a + 1 ⋅ b − 1 b + 1 ⋅ c − 1 c + 1 + a − 1 a + 1 + b − 1 b + 1 + c − 1 c + 1 = a b c − a b − b c − c a + a + b + c − 1 + ∑ ( a − 1 ) ⋅ ( b + 1 ) ⋅ ( c + 1 ) ( a + 1 ) ⋅ ( b + 1 ) ⋅ ( c + 1 ) = a b c − a b − b c − c a + a + b + c − 1 + ∑ ( a b c + a b + a c − b c + a − b − c − 1 ) ( a + 1 ) ⋅ ( b + 1 ) ⋅ ( c + 1 ) = a b c − a b − b c − c a + a + b + c − 1 + 3 a b c + a b + b c + c a − a − b − c − 3 ( a + 1 ) ⋅ ( b + 1 ) ⋅ ( c + 1 ) = 4 a b c − 4 ( a + 1 ) ⋅ ( b + 1 ) ⋅ ( c + 1 ) = 0. ■ \text{证明：}xyz+x+y+z=\frac{a-1}{a+1}\cdot \frac{b-1}{b+1}\cdot \frac{c-1}{c+1}+\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}\\=\frac{abc-ab-bc-ca+a+b+c-1+\sum{\left( a-1 \right) \cdot \left( b+1 \right) \cdot \left( c+1 \right)}}{\left( a+1 \right) \cdot \left( b+1 \right) \cdot \left( c+1 \right)}\\=\frac{abc-ab-bc-ca+a+b+c-1+\sum{\left( abc+ab+ac-bc+a-b-c-1 \right)}}{\left( a+1 \right) \cdot \left( b+1 \right) \cdot \left( c+1 \right)}\\=\frac{abc-ab-bc-ca+a+b+c-1+3abc+ab+bc+ca-a-b-c-3}{\left( a+1 \right) \cdot \left( b+1 \right) \cdot \left( c+1 \right)}\\=\frac{4abc-4}{\left( a+1 \right) \cdot \left( b+1 \right) \cdot \left( c+1 \right)}=0. \blacksquare 证明：xyz+x+y+z=a+1a−1⋅b+1b−1⋅c+1c−1+a+1a−1+b+1b−1+c+1c−1=(a+1)⋅(b+1)⋅(c+1)abc−ab−bc−ca+a+b+c−1+∑(a−1)⋅(b+1)⋅(c+1)=(a+1)⋅(b+1)⋅(c+1)abc−ab−bc−ca+a+b+c−1+∑(abc+ab+ac−bc+a−b−c−1)=(a+1)⋅(b+1)⋅(c+1)abc−ab−bc−ca+a+b+c−1+3abc+ab+bc+ca−a−b−c−3=(a+1)⋅(b+1)⋅(c+1)4abc−4=0.■
想到这一点，题目的难度就大大减小了：
证明：可假设 x = a − 1 a + 1 , y = b − 1 b + 1 , z = c − 1 c + 1 ，有： x y z + x + y + z = 0. 则 L H S − R H S = a + 3 ( a + 1 ) 2 + b + 3 ( b + 1 ) 2 + c + 3 ( c + 1 ) 2 − 3 = 1 2 ⋅ [ ( a − 1 ) 2 − 3 ( a 2 − 1 ) ( a + 1 ) 2 + ( b − 1 ) 2 − 3 ( b 2 − 1 ) ( b + 1 ) 2 + ( c − 1 ) 2 − 3 ( c 2 − 1 ) ( c + 1 ) 2 ] = 1 2 ⋅ ( x 2 + y 2 + z 2 − 3 x − 3 y − 3 z ) = 1 2 ⋅ ( x 2 + y 2 + z 2 + 3 x y z ) ≥ 3 2 x 2 y 2 z 2 3 + 3 2 x y z . \text{证明：可假设}x=\frac{a-1}{a+1}, y=\frac{b-1}{b+1}, z=\frac{c-1}{c+1}\text{，有：}xyz+x+y+z=0.\\\text{则}LHS-RHS=\frac{a+3}{\left( a+1 \right) ^2}+\frac{b+3}{\left( b+1 \right) ^2}+\frac{c+3}{\left( c+1 \right) ^2}-3\\=\frac{1}{2}\cdot \left[ \frac{\left( a-1 \right) ^2-3\left( a^2-1 \right)}{\left( a+1 \right) ^2}+\frac{\left( b-1 \right) ^2-3\left( b^2-1 \right)}{\left( b+1 \right) ^2}+\frac{\left( c-1 \right) ^2-3\left( c^2-1 \right)}{\left( c+1 \right) ^2} \right] \\=\frac{1}{2}\cdot \left( x^2+y^2+z^2-3x-3y-3z \right) \\=\frac{1}{2}\cdot \left( x^2+y^2+z^2+3xyz \right) \ge \frac{3}{2}\sqrt[3]{x^2y^2z^2}+\frac{3}{2}xyz. 证明：可假设x=a+1a−1,y=b+1b−1,z=c+1c−1，有：xyz+x+y+z=0.则LHS−RHS=(a+1)2a+3+(b+1)2b+3+(c+1)2c+3−3=21⋅[(a+1)2(a−1)2−3(a2−1)+(b+1)2(b−1)2−3(b2−1)+(c+1)2(c−1)2−3(c2−1)]=21⋅(x2+y2+z2−3x−3y−3z)=21⋅(x2+y2+z2+3xyz)≥233x2y2z2+23xyz.
又 ∵ x = a − 1 a + 1 , y = b − 1 b + 1 , z = c − 1 c + 1 ∴ ∣ x ∣ < 1 , ∣ y ∣ < 1 , ∣ z ∣ < 1 ∴ ∣ x ∣ 2 3 > ∣ x ∣ , ∣ y ∣ 2 3 > ∣ y ∣ , ∣ z ∣ 2 3 > ∣ z ∣ ∴ a + 3 ( a + 1 ) 2 + b + 3 ( b + 1 ) 2 + c + 3 ( c + 1 ) 2 − 3 ≥ 3 2 ⋅ ( x 2 3 y 2 3 z 2 3 + x y z ) > 3 2 ⋅ ( ∣ x y z ∣ + x y z ) ≥ 0. ■ \text{又}\because x=\frac{a-1}{a+1}, y=\frac{b-1}{b+1}, z=\frac{c-1}{c+1}\\\therefore \left| x \right|<1, \left| y \right|<1, \left| z \right|<1\\\therefore |x|^{\frac{2}{3}}>|x|, |y|^{\frac{2}{3}}>|y|, |z|^{\frac{2}{3}}>|z|\\\therefore \frac{a+3}{\left( a+1 \right) ^2}+\frac{b+3}{\left( b+1 \right) ^2}+\frac{c+3}{\left( c+1 \right) ^2}-3\ge \frac{3}{2}\cdot \left( x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}}+xyz \right) \\>\frac{3}{2}\cdot \left( \left| xyz \right|+xyz \right) \ge 0. \blacksquare 又∵x=a+1a−1,y=b+1b−1,z=c+1c−1∴∣x∣<1,∣y∣<1,∣z∣<1∴∣x∣32>∣x∣,∣y∣32>∣y∣,∣z∣32>∣z∣∴(a+1)2a+3+(b+1)2b+3+(c+1)2c+3−3≥23⋅(x32y32z32+xyz)>23⋅(∣xyz∣+xyz)≥0.■
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