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Difficulty 5.8 AIME, harder Prove it
Example 5 Given that a , b , c a, b, c a , b , c ( a + b ) 3 + 4 c 3 ⩾ 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
(a+b)^{3}+4 c^{3} \geqslant 4\left(\sqrt{a^{3} b^{3}}+\sqrt{b^{3} c^{3}}+\sqrt{c^{3} a^{3}}\right) .
( a + b ) 3 + 4 c 3 ⩾ 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
Solution ( a + b ) 3 + 4 c 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 + 4 c 3 = 2 ( a 2 b + a b 2 ) + ( a 2 + b 2 ) ( a + b ) + 4 c 3 ⩾ 4 a 3 b 3 + ( a 3 2 + b 3 2 ) 2 + 4 c 3 ⩾ 4 a 3 ⋅ b 3 + 4 c 3 2 ( a 3 2 + b 3 2 ) = 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
\begin{array}{l}
(a+b)^{3}+4 c^{3} \\
=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}+4 c^{3} \\
=2\left(a^{2} b+a b^{2}\right)+\left(a^{2}+b^{2}\right)(a+b)+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} b^{3}}+\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right)^{2}+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} \cdot b^{3}}+4 c^{\frac{3}{2}}\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right) \\
=4\left(\sqrt{a^{3} b^{3}}+\sqrt{b^{3} c^{3}}+\sqrt{c^{3} a^{3}}\right) .
\end{array}
( a + b ) 3 + 4 c 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 + 4 c 3 = 2 ( a 2 b + a b 2 ) + ( a 2 + b 2 ) ( a + b ) + 4 c 3 ⩾ 4 a 3 b 3 + ( a 2 3 + b 2 3 ) 2 + 4 c 3 ⩾ 4 a 3 ⋅ b 3 + 4 c 2 3 ( a 2 3 + b 2 3 ) = 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
Prove: By the AM-GM inequality and Cauchy-Schwarz inequality,( a + b ) 3 + 4 c 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 + 4 c 3 = 2 ( a 2 b + a b 2 ) + ( a 2 + b 2 ) ( a + b ) + 4 c 3 ⩾ 4 a 3 b 3 + ( a 3 2 + b 3 2 ) 2 + 4 c 3 ⩾ 4 a 3 ⋅ b 3 + 4 c 3 2 ( a 3 2 + b 3 2 ) = 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
\begin{array}{l}
(a+b)^{3}+4 c^{3} \\
=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}+4 c^{3} \\
=2\left(a^{2} b+a b^{2}\right)+\left(a^{2}+b^{2}\right)(a+b)+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} b^{3}}+\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right)^{2}+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} \cdot b^{3}}+4 c^{\frac{3}{2}}\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right) \\
=4\left(\sqrt{a^{3} b^{3}}+\sqrt{b^{3} c^{3}}+\sqrt{c^{3} a^{3}}\right) .
\end{array}
( a + b ) 3 + 4 c 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 + 4 c 3 = 2 ( a 2 b + a b 2 ) + ( a 2 + b 2 ) ( a + b ) + 4 c 3 ⩾ 4 a 3 b 3 + ( a 2 3 + b 2 3 ) 2 + 4 c 3 ⩾ 4 a 3 ⋅ b 3 + 4 c 2 3 ( a 2 3 + b 2 3 ) = 4 ( a 3 b 3 + b 3 c 3 + c 3 a 3 ) .
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