18.
Áp dụng BĐT quen thuộc: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\) ta có:
\(\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}\ge\dfrac{2}{1+\sqrt{a^3b^3}}\) ; \(\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge\dfrac{2}{1+\sqrt{abc^4}}\)
Cộng vế:
\(\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge2\left(\dfrac{1}{1+\sqrt{a^3b^3}}+\dfrac{1}{1+\sqrt{abc^4}}\right)\ge2\left(\dfrac{2}{1+\sqrt[4]{a^4b^4c^4}}\right)\)
\(\Rightarrow\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge\dfrac{4}{1+abc}\)
\(\Rightarrow\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}\ge\dfrac{3}{1+abc}\) (đpcm)
19.
Biến đổi tương đương:
\(\Leftrightarrow\left(a^2+b^2\right)xy+ab\left(x^2+y^2\right)\ge\left(a^2+b^2+2ab\right)xy\)
\(\Leftrightarrow\left(a^2+b^2\right)xy+ab\left(x^2+y^2\right)\ge\left(a^2+b^2\right)xy+2abxy\)
\(\Leftrightarrow ab\left(x^2+y^2\right)-2abxy\ge0\)
\(\Leftrightarrow ab\left(x^2+y^2-2xy\right)\ge0\)
\(\Leftrightarrow ab\left(x-y\right)^2\ge0\)
20.
\(\Leftrightarrow a+b+c\ge\dfrac{3\left(ab+bc+ca\right)}{abc}\)
\(\Leftrightarrow a+b+c\ge\dfrac{3\left(ab+bc+ca\right)}{a+b+c}\) (do \(abc=a+b+c\))
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
21.
Ta có:
\(a^3+ab^2\ge2a^2b\)
\(a^3+ac^2\ge2a^2c\)
\(b^3+a^2b\ge2ab^2\)
\(b^3+c^2b\ge2b^2c\)
\(c^3+a^2c\ge2ac^2\)
\(c^3+b^2c\ge2bc^2\)
Cộng vế với vế:
\(2\left(a^3+b^3+c^3\right)+a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)\ge2a^2\left(b+c\right)+2b^2\left(c+a\right)+2c^2\left(a+b\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)\)
22.
\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow2\left(ab\right)^2+2\left(bc\right)^2+2\left(ca\right)^2-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow\left(ab-bc\right)^2+\left(ab-ca\right)^2+\left(bc-ca\right)^2\ge0\)
23.a.
\(\Leftrightarrow\left(a+b\right)\left(a^2+b^2\right)\le2\left(a^3+b^3\right)\)
\(\Leftrightarrow a^3+b^3+a^2b+ab^2\le2\left(a^3+b^3\right)\)
\(\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) (luôn đúng với a;b;c không âm)
23b.
Áp dụng câu a: \(\dfrac{a+b}{2}.\dfrac{a^2+b^2}{2}\le\dfrac{a^3+b^3}{2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{a^3+b^3}{2}.\dfrac{a^4+b^4}{2}\le\dfrac{a^7+b^7}{2}\)
\(\Leftrightarrow\left(a^3+b^3\right)\left(a^4+b^4\right)\le2\left(a^7+b^7\right)\)
\(\Leftrightarrow a^7-a^4b^3+b^7-a^3b^4\ge0\)
\(\Leftrightarrow a^4\left(a^3-b^3\right)-b^4\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a^4-b^4\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2\right)\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
23c.
Ta sẽ chứng minh: \(\dfrac{a+b}{2}.\dfrac{a^3+b^3}{2}\le\dfrac{a^4+b^4}{2}\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
Nên ta chỉ cần chứng minh: \(\dfrac{a^4+b^4}{2}.\dfrac{a^5+b^5}{2}\le\dfrac{a^9+b^9}{2}\)
\(\Leftrightarrow\left(a^4+b^4\right)\left(a^5+b^5\right)\le2\left(a^9+b^9\right)\)
\(\Leftrightarrow a^9-a^5b^4+b^9-a^4b^5\ge0\)
\(\Leftrightarrow\left(a^4-b^4\right)\left(a^5-b^5\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\ge0\) (luôn đúng)
24.
\(\dfrac{\left(a+b\right)\left(1-ab\right)}{\left(1+a^2\right)\left(1+b^2\right)}\ge-\dfrac{1}{2}\)
\(\Leftrightarrow\left(1+a^2\right)\left(1+b^2\right)+2\left(a+b\right)\left(1-ab\right)\ge0\)
\(\Leftrightarrow a^2+b^2+a^2b^2+1-2ab\left(a+b\right)+2\left(a+b\right)\ge0\)
\(\Leftrightarrow a^2+b^2+2ab+a^2b^2-2ab+1-2ab\left(a+b\right)+2\left(a+b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)+1-2ab\left(a+b+1\right)+a^2b^2\ge0\)
\(\Leftrightarrow\left(a+b+1\right)^2-2ab\left(a+b+1\right)+a^2b^2\ge0\)
\(\Leftrightarrow\left(a+b+1-ab\right)^2\ge0\) (luôn đúng)
Tương tự: \(\dfrac{\left(a+b\right)\left(1-ab\right)}{\left(1+a^2\right)\left(1+b^2\right)}\le\dfrac{1}{2}\Leftrightarrow\left(ab+a+b-1\right)^2\ge0\) (luôn đúng)
25.
Ta có:
\(a^3b+abc^2\ge2a^2bc\)
\(b^3c+a^2bc\ge2ab^2c\)
\(c^3a+ab^2c\ge2abc^2\)
Cộng vế với vế:
\(a^3b+b^3c+c^3a+abc\left(a+b+c\right)\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)
26.
Ta chứng minh BĐT phụ sau:
Với mọi x;y dương ta luôn có: \(x^3+y^3\ge xy\left(x+y\right)\)
Thật vậy, BĐT tương đương:
\(x^3-x^2y+y^3-xy^2\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (luôn đúng)
Áp dụng:
\(VT=\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ca}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ca}\)
\(VT\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\) (đpcm)
