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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 \(a=b=c\)
bài 1. ta có
\(a^2+b^2+c^2+d^2\ge ab+ac+ad\)
\(\Leftrightarrow b^2+ab+\frac{a^2}{4}+c^2+ac+\frac{a^2}{4}+d^2+ad+\frac{a^2}{4}+\frac{a^2}{4}\ge0\)
\(\Leftrightarrow\left(b+\frac{a}{2}\right)^2+\left(c+\frac{a}{2}\right)^2+\left(d+\frac{a}{2}\right)^2+\frac{a^2}{4}\ge0\) luôn đúng
Bài 2
ta có \(\frac{a^5}{b^5}+1+1+1+1\ge\frac{5.a}{b}\) (bất đẳng thức cauchy)
Tương tự ta có \(\frac{b^5}{c^5}+4\ge\frac{5b}{c};\frac{c^5}{a^5}+4\ge\frac{5c}{a}\)
\(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\)
Mà dễ dàng chứng minh \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\)
Nên ta có \(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
bài 1 : \(^{a^2+B^2+C^2+D^2}\)>hoặc =ab+ac+ad
\(^{a^2+b^2+c^2}\)- ab-ac-ad>hoặc = 0
\((\frac{1}{4}^{a^2-ab+b^2})+(\frac{1}{4}^{a^2-ac+c^2})+(\frac{1}{4}^{a^2-ad+d^2})\)>hoặc =0
\((\frac{1}{2}a-b)^2+(\frac{1}{2}a-c)^2+(\frac{1}{2}a-d)^2>=0\)
Vì \((\frac{1}{2}a-b)^2>=0\)với mọi \(A,b\varepsilon n\)
=> đpcm tự kết luận
a/ Biến đổi tương đương:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh
b/ \(VT=\frac{a-d}{b+d}+1+\frac{d-b}{b+c}+1+\frac{b-c}{a+c}+1+\frac{c-a}{a+d}+1-4\)
\(VT=\frac{a+b}{b+d}+\frac{c+d}{b+c}+\frac{a+b}{a+c}+\frac{c+d}{a+d}-4\)
\(VT=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)
\(\Rightarrow VT\ge\left(a+b\right).\frac{4}{b+d+a+c}+\left(c+d\right).\frac{4}{b+c+a+d}-4\)
\(\Rightarrow VT\ge\frac{4}{\left(a+b+c+d\right)}\left(a+b+c+d\right)-4=4-4=0\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
\(\frac{a^4}{a^3+2b^3}=a-\frac{2ab^3}{a^3+b^3+b^3}\ge a-\frac{2ab^3}{3\sqrt[3]{a^3.b^3.b^3}}=a-\frac{2}{3}b\)
Tương tự ta có
\(\frac{b^4}{b^3+2c^3}\ge b-\frac{2}{3}c\) ; \(\frac{c^4}{c^3+2d^3}\ge c-\frac{2}{3}d\) ; \(\frac{d^4}{d^3+2a^3}\ge d-\frac{2}{3}a\)
Cộng vế với vế:
\(VT\ge a+b+c+d-\frac{2}{3}\left(a+b+c+d\right)=\frac{a+b+c+d}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}=\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\)
Áp dụng BĐT Cauchy-Schwar dạng Engel ta có:
\(\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\ge\frac{\left(1+1+2+4\right)^2}{a+b+c+d}\)
\(=\frac{8^2}{a+b+c+d}=\frac{64}{a+b+c+d}=VP\)