Bài bất đẳng thức hay

Posted on April 18, 2014 by cuoichutdi

Bài toán: Cho các số dương a,b,c,d thoả \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}=4

Chứng minh rằng:

\sqrt[3]{\dfrac{a^3+b^3}{2}}+\sqrt[3]{\dfrac{b^3+c^3}{2}}+\sqrt[3]{\dfrac{c^3+d^3}{2}}+\sqrt[3]{\dfrac{d^3+a^3}{2}} \leq 2(a+b+c+d-2)

Lời giải:

Ta chứng minh đánh giá:

\sqrt[3]{\dfrac{a^3+b^3}{2}} \leq \dfrac{a^2+b^2}{a+b}.

Tương đương với: (a+b)^3(a^3+b^3) \leq 2(a^2+b^2)^3.

hay (a-b)^4(a^2+ab+b^2) \geq 0.

Do đó ta cần chứng minh:

\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+d^2}{c+d}+\dfrac{d^2+a^2}{d+a} \leq 2(a+b+c+d-2)

Chú ý: \dfrac{a^2+b^2}{a+b}=a+b-\dfrac{2ab}{a+b}.Nên ta cần chứng minh:

2 \leq \dfrac{ab}{a+b}+\dfrac{cb}{c+b}+\dfrac{cd}{c+d}+\dfrac{ad}{a+d}

Mà theo Cauchy Schwarz ta có:

\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}}+\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{d}} \geq 2.

Vậy ta có:\sqrt[3]{\dfrac{a^3+b^3}{2}}+\sqrt[3]{\dfrac{b^3+c^3}{2}}+\sqrt[3]{\dfrac{c^3+d^3}{2}}+\sqrt[3]{\dfrac{d^3+a^3}{2}} \leq 2(a+b+c+d-2)

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