Ketaksamaan jangan dibongkar

Soal. Jika a,b,c,d bilangan real positif sehingga a+b+c+d=4, tunjukkan bahwa

\dfrac{4}{abcd} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}.

Bukti. Ketaksamaan dapat ditulis menjadi

4 \ge a^2cd+b^2da+c^2ab+d^2bc.

Misalkan p,q,r,s adalah suatu permutasi dari a,b,c,d, dan misalkan p \ge q \ge r \ge s. Perhatikan bahwa pqr \ge pqs \ge prs \ge qrs. Perhatikan pula bahwa p+q+r+s=4.

Perhatikan bahwa dengan ketaksamaan re-arrangement, berlaku

a \cdotp acd + b \cdotp bda + c \cdotp cab + d \cdotp dbc \le p \cdotp pqr + q \cdotp pqs + r \cdotp prs + s \cdotp qrs

=p^2qr+pq^2s+pr^2s+qrs^2=pq(pr+qs)+rs(pr+qs)=(pq+rs)(pr+qs)

\le \dfrac{1}{4}(pq+rs+pr+qs)^2 =\dfrac{1}{4}((p+s)(q+r))^2

\le \dfrac{1}{64}(p+s+q+r)^4 = 4.

Terbukti.

Refleksi.

 

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