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Bài 1:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)
\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)
\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Bài 2:
\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))
\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)
chị giải thích cho em cái đoạn này với ạ
\(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Bài 1:ta có BĐt \(a^3+b^3\ge ab\left(a+b\right)\)vì nó tương đương với \(\left(a+b\right)\left(a-b\right)^2\ge0\)(luôn đúng với a,b>0)
Áp dụng vào bài toán:
\(\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ac}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ac}=a+b+c\)dấu = xảy ra khi a=b=c
bài 2:
cần chứng minh \(\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge0\)
hay \(\dfrac{a-b}{b+c}+1+\dfrac{b-c}{c+d}+1+\dfrac{c-d}{d+a}+1+\dfrac{d-a}{a+b}+1\ge4\)
\(\Leftrightarrow\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}\ge4\)
xét \(VT=\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)+\left(b+d\right)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)\)
Áp dụng BĐT cauchy dạng phân thức:
\(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d};\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{a+b+c+d}\)
do đó \(VT\ge\dfrac{4\left(a+c\right)}{a+b+c+d}+\dfrac{4\left(b+d\right)}{a+b+c+d}=4\)
dấu = xảy ra khi a=b=c=d
Từ bài toán này (mà bạn đã hỏi cách đây vài bữa):
cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\) - Hoc24
Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)
Do đó: \(VT\ge\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)
Lại có: \(\dfrac{a+b+c}{\sqrt[3]{abc}}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3\)
Đặt \(\dfrac{a+b+c}{\sqrt[3]{abc}}=x\ge3\Rightarrow VT\ge x+\dfrac{1}{x}=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\ge2\sqrt{\dfrac{x}{9x}}+\dfrac{8}{9}.3=\dfrac{10}{3}\) (đpcm)
Áp dụng BĐT \(AM-GM\) ta có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{15}b^4}{b^9}}=5\dfrac{a^3}{b}\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{15}c^4}{c^9}}=5\dfrac{b^3}{c}\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{15}a^4}{a^9}}=5\dfrac{c^3}{a}\)
Cộng từng vế của BĐT ta được :
\(3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
Tiếp tục áp dụng BĐT \(AM-GM\) ta lại có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{10}b^6}{b^6}}=5a^2\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{10}c^6}{c^6}}=5b^2\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{10}a^6}{a^6}}=5c^2\)
Cộng vế theo vế ta được :
\(2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+3\left(a^2+b^2+c^2\right)\ge5\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow5\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\left(đpcm\right)\)
Bài 3:
Áp dụng bất đẳng thức AM - GM có:
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}\)
\(=2+2+2=6\)
Dấu " = " khi x = y = z = 1
Vậy...
3. Với x,y,z>0 áp dụng BĐT Cauchy ta có
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)\)
\(\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}=2+2+2=6\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\Leftrightarrow x=y=z=1\)
1. Với a=b=c=0, ta thấy BĐT trên đúng
Với a,b,c>0 áp dụng BĐT Cauchy cho 3 số dương
\(a^3+a^3+b^3\ge3\sqrt[3]{a^3.a^3.b^3}=3\sqrt[3]{a^6b^3}=3a^2b\) (1)
\(b^3+b^3+c^3\ge3\sqrt[3]{b^3.b^3.c^3}=3\sqrt[3]{b^6c^3}=3b^2c\) (2)
\(c^3+c^3+a^3\ge3\sqrt[3]{c^3.c^3.a^3}=3\sqrt[3]{c^6a^3}=3c^2a\) (3)
Cộng (1), (2), (3) vế theo vế:
\(a^3+b^3+c^3\ge a^2b+b^2c+c^2a>\dfrac{a^2b+b^2c+c^2a}{3}\) (vì a,b,c>0)
Do đó BĐT trên đúng \(\forall a,b,c\ge0\)
Ta có \(\dfrac{a^3+b^3}{2ab}\ge\dfrac{ab\left(a+b\right)}{2ab}=\dfrac{a+b}{2}\)
(áp dụng BĐT quen thuộc \(a^3+b^3\ge ab\left(a+b\right)\))
Lập 2 BĐT tương tự rồi cộng theo vế:
\(VT\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có đpcm.