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BĐT cần chứng minh tương đương :
\(\sqrt{\dfrac{a^2+b^2}{2}}-\sqrt{ab}\ge\dfrac{a+b}{2}-\dfrac{2ab}{a+b}\)
\(\Leftrightarrow\dfrac{\dfrac{a^2+b^2}{2}-ab}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a+b\right)^2-4ab}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\right)\ge0\)
ta phải chứng minh;
\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\)\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{1}{2\left(a+b\right)}\)
\(\Leftrightarrow a+b\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\)\(\Leftrightarrow2a+2b-\sqrt{2\left(a^2+b^2\right)}-2\sqrt{ab}\ge0\)
\(\Leftrightarrow\left(a+b-\sqrt{2\left(a^2+b^2\right)}\right)+\left(a+b-2\sqrt{ab}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2-2\left(a^2+b^2\right)}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a+b\right)^2-4ab}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\dfrac{-\left(a-b\right)^2}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a-b\right)^2}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right)\ge0\)
ta phải chứng minh
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\ge0\)
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}\ge\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\)
\(\Leftrightarrow a+b+2\sqrt{ab}\le a+b+\sqrt{2\left(a^2+b^2\right)}\)
\(\Leftrightarrow2\sqrt{ab}\le\sqrt{2\left(a^2+b^2\right)}\Leftrightarrow\left(a-b\right)^2\ge0\)
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
Thay \(a=b=c=0,25\)thì ta có:
\(\dfrac{1}{\sqrt{0,25}}+\dfrac{1}{\sqrt{0,25}}+\dfrac{2\sqrt{2}}{\sqrt{0,25}}\approx9,657\)
\(\dfrac{8}{0,25+0,25+0,25}\approx10,667\)
Vậy đề sai
1) Đặt T là vế trái của BĐT
Áp dụng BĐT Cauchy-Schwarz và AM-GM, ta có:
\(T=\dfrac{x^4}{xy}+\dfrac{y^4}{yz}+\dfrac{z^4}{xz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+xz}\ge\dfrac{1}{x^2+y^2+z^2}=1\)
Vậy ta có đpcm.Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
3)b) Đặt T là vế trái, áp dụng AM-GM ta có:
\(b+c=\left(b+c\right)\left(a+b+c\right)^2\ge\left(b+c\right)4a\left(b+c\right)=4a\left(b+c\right)^2\ge16abc\)
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
Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:
\(VT=\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{3abc}{2abc}+\dfrac{2ab}{c^2+ab}+\dfrac{2bc}{a^2+bc}+\dfrac{2ac}{b^2+ac}=\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\)
Áp dụng bất đẳng thức \(Cauchy-Schwarz\) \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}=\dfrac{a^2b^2}{c^2ab+a^2b^2}+\dfrac{b^2c^2}{a^2bc+b^2c^2}+\dfrac{a^2c^2}{b^2ac+a^2c^2}\ge\dfrac{\left(ab+bc+ac\right)^2}{c^2ab+a^2b^2+a^2bc+b^2c^2+b^2ac+a^2c^2}\)
Đặt: \(\left\{{}\begin{matrix}ab=x\\bc=y\\ac=z\end{matrix}\right.\) ta được: \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+xy+xz+xy}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)
Nên: \(\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)
Mà: \(VT\ge\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\Leftrightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Lời giải:
Áp dụng BĐT AM-GM ta có: \(\frac{a^3+b^3+c^3}{2abc}\geq \frac{3\sqrt[3]{a^3b^3c^3}}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\) (1)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{a^2+b^2+c^2+ab+bc+ac}\) (2)
Có:
\((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2=2(a^2+b^2+c^2)+2\sqrt{(a^2+b^2)(b^2+c^2)}+2\sqrt{(b^2+c^2)(c^2+a^2)}+\sqrt{(a^2+b^2)(c^2+a^2)}\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(a^2+b^2)(b^2+c^2)}\geq \sqrt{(ac+b^2)^2}=ac+b^2\)
\(\sqrt{(b^2+c^2)(c^2+a^2)}\geq \sqrt{(ba+c^2)^2}=ba+c^2\)
\(\sqrt{(a^2+b^2)(c^2+a^2)}\geq \sqrt{(a^2+bc)^2}=a^2+bc\)
\(\Rightarrow (\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 2(a^2+b^2+c^2)+2(a^2+b^2+c^2+ab+bc+ac)\)
\(\geq a^2+b^2+c^2+ab+bc+ac+2(a^2+b^2+c^2+ab+bc+ac)\) (AM-GM)
Hay \((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 3(a^2+b^2+c^2+ab+bc+ac)\) (3)
Từ \((2); (3)\Rightarrow \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq 3\) (4)
Từ \((1); (4)\Rightarrow \frac{a^3+b^3+c^3}{2abc}+\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ac}\geq \frac{9}{2}\)
Ta có đpcm.
Dấu bằng xảy ra khi $a=b=c$
Ủa bài này hỏi rồi hỏi gì nữa?
Chắc là do em nhớ nhầm ạ