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Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
a) \(\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\left(đpcm\right)\)
Áp dụng BĐT Cô -si cho 3 số dương:
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Ta có: \(\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right)\)
\(=\left(ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\right)\left(c+\frac{1}{c}\right)\)
\(=\left[ab+\frac{1}{16ab}+\frac{15}{16ab}+\left(\frac{a}{b}+\frac{b}{a}\right)\right]\left(c+\frac{1}{c}\right)\)
\(\ge\left[2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{4\left(a+b\right)^2}+2\sqrt{\frac{a}{b}.\frac{b}{a}}\right]\left(2\sqrt{c.\frac{1}{c}}\right)\)
\(\ge\frac{25}{2}\left(Đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2};c=1\)
Xí trước phần b
Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2b+ca^2}+\frac{ca}{b^2c+ab^2}+\frac{ab}{c^2a+bc^2}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}+\frac{c^2a^2}{ab^2c^2+a^2b^2c}+\frac{a^2b^2}{a^2bc^2+ab^2c^2}\)
\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cách làm khác của phần b ngắn gọn hơn:)
Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
Đặt ⎧⎪⎨⎪⎩a+b−c=xb+c−a=yc+a−b=z(x,y,z>0){a+b−c=xb+c−a=yc+a−b=z(x,y,z>0)
⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a=z+x2b=x+y2c=y+z2⇒{a=z+x2b=x+y2c=y+z2
⇒√a(1b+c−a−1√bc)=√2(z+x)2(1y−2√(x+y)(y+z))≥√x+√z2(1y−2√xy+√yz)=√x+√z2y−1√y⇒a(1b+c−a−1bc)=2(z+x)2(1y−2(x+y)(y+z))≥x+z2(1y−2xy+yz)=x+z2y−1y
Tương tự
⇒∑√a(1b+c−a−1√bc)≥∑√x+√z2y−∑1√y⇒∑a(1b+c−a−1bc)≥∑x+z2y−∑1y
⇒VT≥∑[x√x(y+z)]2xyz−∑√xy√xyz≥2√xyz(x+y+z)2xyz−x+y+z√xyz≐x+y+z√xyz−x+y+z√xyz=0⇒VT≥∑[xx(y+z)]2xyz−∑xyxyz≥2xyz(x+y+z)2xyz−x+y+zxyz≐x+y+zxyz−x+y+zxyz=0
(∑√xy≤x+y+z,x√x(y+z)≥2x√xyz)(∑xy≤x+y+z,xx(y+z)≥2xxyz)
dấu = ⇔x=y=z⇔a=b=c
dễ mà SD BDT cô-si
áp dụng BĐT cosi là ra