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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
\(2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2\left(a^2+b^2+c^2\right)+4\frac{ab+bc+ca}{abc}.\)
\(=2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)(vì abc=1)
\(=2\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\)
\(=2\left(a+b+c\right)^2\)
Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)(bất đẳng thức cô si cho ba số không âm)
Đặt \(a+b+c=x\ge3\)
Dễ thấy : \(2x^2-7x+3=\left(2x-1\right)\left(x-3\right)\ge0\)
Hay \(2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)
Dấu '=' xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=3\end{cases}\Leftrightarrow}a=b=c=1\)
Đặt A = a + b + c .
Áp dụng BĐT Cosi cho 3 số thực dương ta có : \(A\ge3^3\sqrt{abc}=3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\cdot\frac{ab+bc+ca}{abc}-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2A^2-7A+3=\left(2A-1\right)\left(A-3\right)\ge0\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\left(a+b+c\right)\)(\(\frac{1}{a}\)\(+\)\(\frac{1}{b}\)\(+\)\(\frac{1}{c}\))\(=\)\(1+\frac{a}{b}\)\(+\)\(\frac{a}{c}\)\(+1\)\(\frac{b}{c}\)\(+\)\(\frac{b}{a}\)\(+1\)\(+\frac{c}{b}\)\(+\frac{c}{a}\)
\(=\)\(3\)\(+\)(\(\frac{a}{b}\)\(+\frac{b}{a}\))\(+\)\(\frac{c}{b}\)\(+\)\(\frac{b}{c}\))\(+\)(\(\frac{a}{c}\)\(+\)\(\frac{c}{a}\))
\(mà\)\(\frac{a}{b}\)\(+ \)\(\frac{b}{a}\)\(>=2\)\(;\)\(\frac{b}{c}\)\(+\)\(\frac{c}{b}\)\(>=2\)\(;\)\(\frac{a}{c}\)\(+\)\(\frac{c}{a}\)\(>=2\)( cái này bạn tự chứng minh được)
\(=>\)\(\left(a+b+c\right)\)(\(\frac{1}{a}\)\(+\)\(\frac{1}{b}\)\(+\)\(\frac{1}{c}\)) \(>=3+2+2+2\)
\(=>\)\(\left(a+b+c\right)\)(\(\frac{1}{a}\)\(+\)\(\frac{1}{b}\)\(+\)\(\frac{1}{c}\)) \(>=9\)(\(luôn\)\(đúng\)\(với\)\(mọi\)\(a,b,c\)\(dương\))
\(k\)\(cho\)\(mình\)\(nha\)\(các\)\(bạn\), \(mình\)\(k\)\(lại\)\(cho\)\(nhé\)
\(chúc\)\(các\)\(bạn\)\(học\)\(tốt\)
Áp dụng Cachy cho 3 số ra ngay kết quả em nhé!
hoặc cách 2: ÁP dụng BUN cho 3 số
\(\left(\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2\right)\left(\frac{1}{\sqrt{a}^2}+\frac{1}{\sqrt{b}^2}+\frac{1}{\sqrt{c}^2}\right)\ge\)
\(\left(\sqrt{a}.\frac{1}{\sqrt{a}}+\sqrt{b}.\frac{1}{\sqrt{b}}+\sqrt{c}.\frac{1}{\sqrt{c}}\right)^2=3^2=9\)
+ Áp dụng bđt AM-GM cho 3 số dương a,b và c ta có :
\(a+b+c\ge3\sqrt[3]{abc}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
+ Áp dụng bđt AM-GM cho 3 số dương \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
Do đó : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc\cdot\frac{1}{abc}}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Đặt A= \(\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=1+\frac{b}{a}+\frac{c}{a}+1+\frac{a}{b}+\frac{c}{b}+1+\frac{b}{c}+\frac{a}{c}\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)
\(\ge2\) \(\ge2\) \(\ge2\)
=>\(A\ge9\left(đpcm\right)\)