Let \(A=\dfrac{a^3+b^3+c^3+d^3}{a+b+c+d}\)

\(A=\dfrac{a^3}{a+b+c+d}+\dfrac{b^3}{a+b+c+d}+\dfrac{c^3}{a+b+c+d}+\dfrac{d^3}{a+b+c+d}\)

\(A=\dfrac{a^4}{a^2+ab+ac+ad}+\dfrac{b^4}{ab+b^2+bc+bd}+\dfrac{c^4}{ac+bc+c^2+cd}+\dfrac{d^4}{ad+bd+cd+d^2}\)

Following Schwarz's inequality , we have :

\(A\ge\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)}=\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{\left(a+b+c+d\right)^2}\)

Following Bunyakovsky's inequality , we have :

\(\left(a^2+b^2+c^2+d^2\right)\left(1+1+1+1\right)\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow4\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow A\ge\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{4\left(a^2+b^2+c^2+d^2\right)}=\dfrac{a^2+b^2+c^2+d^2}{4}\ge\dfrac{4\sqrt[4]{a^2b^2c^2d^2}}{4}=\dfrac{4.1}{4}=1\)

\(\Rightarrow a^3+b^3+c^3+d^3\ge a+b+c+d\)

Equality occurs when :

a = b = c = d = 1

Let A=a3+b3+c3+d3a+b+c+d

A=a3a+b+c+d+b3a+b+c+d+c3a+b+c+d+d3a+b+c+d

A=a4a2+ab+ac+ad+b4ab+b2+bc+bd+c4ac+bc+c2+cd+d4ad+bd+cd+d2

Following Schwarz's inequality , we have :

A≥(a2+b2+c2+d2)2a2+b2+c2+d2+2(ab+ac+ad+bc+bd+cd)=(a2+b2+c2+d2)2(a+b+c+d)2

Following Bunyakovsky's inequality , we have :

(a2+b2+c2+d2)(1+1+1+1)≥(a+b+c+d)2

⇒4(a2+b2+c2+d2)≥(a+b+c+d)2

⇒A≥(a2+b2+c2+d2)24(a2+b2+c2+d2)=a2+b2+c2+d24≥44√a2b2c2d24=4.14=1

⇒a3+b3+c3+d3≥a+b+c+d

Equality occurs when :

a = b = c = d = 1

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