Lời giải.
Áp dụng bất đẳng thức Bunyakovsky dạng phân thức, ta được: $\frac{a^5+b^5+c^5+d^5}{a+b+c+d}=\frac{\frac{a^6}{a}+\frac{b^6}{b}+\frac{c^6}{c}+\frac{d^6}{d}}{a+b+c+d}\geqslant \frac{\frac{(a^3+b^3+c^3+d^3)^2}{a+b+c+d}}{a+b+c+d}=\frac{(a^3+b^3+c^3+d^3)^2}{(a+b+c+d)^2}=\frac{(\frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}+\frac{d^4}{d})^2}{(a+b+c+d)^2}\geqslant \frac{[\frac{(a^2+b^2+c^2+d^2)^2}{a+b+c+d}]^2}{(a+b+c+d)^2}=(\frac{a^2+b^2+c^2+d^2}{a+b+c+d})^4\geqslant (\frac{(a+b+c+d)^2}{4(a+b+c+d)})^4=(\frac{a+b+c+d}{4})^4\geqslant abcd$
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