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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
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
Chứng minh bất đẳng thức \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Có: \(\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2\right)\ge\left(a+b+c\right)^2\) (Bunyakovsky)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
abc = 1 => a^2.b^2.c^2 = 1
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{a^2b^2c^2}{c^3\left(a+b\right)}\)
\(=\frac{\left(bc\right)^2}{ab+ac}+\frac{\left(ac\right)^2}{bc+ba}+\frac{\left(ab\right)^2}{ca+cb}\ge\frac{\left(ab+ac+bc\right)^2}{2\left(ab+ac+bc\right)}=\frac{\left(ab+ac+bc\right)}{2}\)
\(\ge\frac{3\sqrt[3]{ab.ac.bc}}{2}\)(Cauchy) \(=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\\frac{bc}{ab+ac}=\frac{ac}{bc+ba}+\frac{ab}{ca+cb}\Leftrightarrow\end{cases}a=b=c}\)
Mà abc=1 <=> a^3 = 1 <=> a=1 => b=c=a=1
https://diendantoanhoc.net/topic/80159-ch%E1%BB%A9ng-minh-frac1a2b3cfrac12a3bcfrac13bb2c-leqslant-frac316/
bạn tham khảo ở đây nhé
\(P=\frac{a^2}{b^2+2bc}+\frac{b^2}{c^2+2ac}+\frac{c^2}{a^2+2ab}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" xảy ra khi \(a=b=c\)
Giả sử:
\(a>b>c\Rightarrow a-b>0,b-c>0,a-c>0\)
Ta có:
\(\hept{\begin{cases}a^2+b^2+c^2\ge a^2+c^2\\\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}\ge\frac{\left(\frac{1}{a-b}+\frac{1}{b-c}\right)^2}{2}\ge\frac{8}{\left(a-c\right)^2}\end{cases}}\)
Từ đây ta có:
\(VT\ge\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\)
Ta chứng minh
\(\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+c\right)^2\ge0\)(Đúng)
Vậy ta có điều phải chứng minh là đúng. Dấu = xảy ra khi a = - c; b = 0 và các hoán vị của nó
Áp dụng BĐT C-S ta có:
\(\left(a+b+c\right)\cdot VT\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\left(1\right)\)
Mặt khác ta cũng có bổ đề \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\frac{9}{4}\left(2\right)\)
Từ (1) và (2) ta dc DPCM
Dấu "=" xay ra khi \(a=b=c\)
Ta có \(\left(b+c\right)^2\le2\left(b^2+c^2\right)\)
=> \(\frac{a^2}{a^2+\left(b+c\right)^2}\ge\frac{a^2}{a^2+2b^2+c^2}\)
=> \(VT\ge\Sigma\frac{a^4}{a^4+2b^2a^2+2a^2c^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+4\left(a^2b^2+b^2c^2+c^2a^2\right)}\)
=> \(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+2\left(a^2b^2+b^2c^2+a^2c^2\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+\frac{2}{3}\left(a^2+b^2+c^2\right)^2}=\frac{3}{5}\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
Đặt \(x^2+2y^2=m;y^2+2z^2=n;z^2+2x^2=p\)
Ta có :\(9\left(x^2+y^2+z^2\right)\left(\frac{a^3}{x^2+2y^2}+\frac{b^3}{y^2+2z^2}+\frac{c^3}{z^2+2x^2}\right)\)
\(=\left(1+1+1\right)\left(m+n+p\right)\left(\frac{a^3}{m}+\frac{b^3}{n}+\frac{c^3}{p}\right)\ge\left(a+b+c\right)^3=1\)
do đó \(9\left(x^2+y^2+z^2\right)\left(\frac{a^3}{x^2+2y^2}+\frac{b^3}{y^2+2z^2}+\frac{c^3}{z^2+2x^2}\right)\ge1\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(\frac{a^3}{x^2+2y^2}+\frac{b^3}{y^2+2z^2}+\frac{c^3}{z^2+2x^2}\right)\ge\frac{1}{9}\)(đpcm)
Xong rồi đấy,bạn k cho mình nhé
dùng bất đẳng thức svacxơ nha bạn