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1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Lần sau đăng ít một thôi toàn bài dài :v, ko phải ko làm mà là ngại làm
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{2a+b+c}=\frac{a}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{a+2b+c}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right);\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{4}\)
Xảy ra khi \(a=b=c\)
b)Đặt \(THANG=abc\left(a^2+bc\right)\left(b^2+ac\right)\left(c^2+ab\right)>0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b+c}{a^2+bc}-\frac{c+a}{b^2+ac}-\frac{a+b}{a^2+ab}\)
\(=\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4c^2a^2-c^4a^2b^2}{THANG}\)
\(=\frac{\left(a^2b^2-b^2c^2\right)^2+\left(b^2c^2-c^2a^2\right)+\left(c^2a^2-a^2b^2\right)^2}{2THANG}\ge0\) (Đúng)
Xảy ra khi \(a=b=c\)
c)Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)
Và \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ab\left(b-a\right)}{\left(c+a\right)\left(c^2+a^2\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(b+a\right)\left(b^2+a^2\right)}\)
Cộng theo vế 3 đăng thức trên ta có:
\(VT-VP=Σ\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)
\(=\left(a^2+b^2+c^2+ab+bc+ca\right)\cdotΣ\frac{ab\left(a-b\right)^2}{\left(b+c\right)\left(c+a\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\ge0\)
2 bài cuối full quy đồng mệt thật :v
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
<=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2c}{abc}+\frac{2b}{abc}+\frac{2a}{abc}=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2a+2b+2c}{abc}=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2abc}{abc}=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+2=4\)
<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2^2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=2^2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{abc}{abc}=2^2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
đpcm
Bài 1: diendantoanhoc.net
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành
\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)
\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)
Theo BĐT AM-GM và Cauchy-Schwarz ta có:
\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)
Bổ sung bài 1:
BĐT được chứng minh
Đẳng thức xảy ra <=> a=b=c
Đặt \(a+b+c=t\) ta có \(a+b+c\le3\)
Đặt \(P=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow2P\ge\frac{18}{a+b+c}+3\left(a+b+c\right)=\frac{18}{t}+3t\)
ĐẾn đây nhóm thế nào hả ad
Do \(a;b;c>0\) và \(a^2+b^2+c^2=3\)
\(\Rightarrow0< a;b;c< \sqrt{3}\)
Ta cần CM: \(\frac{1}{a}+\frac{3}{2}a\ge\frac{a^2+9}{4}\)
Hay \(\frac{\left(a-1\right)^2\left(4-a\right)}{4a}\ge0\) Dúng do \(0>a< \sqrt{3}\)
Tương tự cộng lại ta được BđT cần cm
Đặt \(x=b+c;\) \(y=a+c;\) và \(z=a+b\)
thì \(a=\frac{y+z-x}{2};\) \(b=\frac{x+z-y}{2};\) và \(c=\frac{x+y-z}{2}\)
Khi đó, bất đẳng trên được quy về dạng sau:
\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}+\frac{x}{2y}+\frac{z}{2y}-\frac{1}{2}+\frac{x}{2z}+\frac{y}{2z}-\frac{1}{2}\)
\(=\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)-\frac{3}{2}\)
Áp dụng BTĐ Cauchy đối với từng phân thức ở vế phải (với \(x,y,z\) là các số dương), ta được
\(\frac{y}{2x}+\frac{x}{2y}\ge2\sqrt{\frac{y}{2x}.\frac{x}{2y}}=1\)
\(\frac{z}{2x}+\frac{x}{2z}\ge2\sqrt{\frac{z}{2x}.\frac{x}{2z}}=1\)
\(\frac{z}{2y}+\frac{y}{2z}\ge2\sqrt{\frac{z}{2y}.\frac{y}{2z}}=1\)
nên \(\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)\ge1+1+1=3\)
Do đó, \(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\ge3-\frac{3}{2}=\frac{3}{2}\)
Dấu \("="\) xảy ra xảy ra khi và chỉ khi \(a=b=c\)
Ta có: \(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)
\(\Leftrightarrow F=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)
\(\Leftrightarrow F\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=P\)
\(\Leftrightarrow P=\frac{a^2+b^2+c^2+2ab+2bc+2cd+2ad+2ac+2bd}{ab+ac+bc+bd+cd+ac+ad+bd}\)
\(\Leftrightarrow P=\frac{\left(a^2+c^2\right)+\left(b^2+d^2\right)+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+bc+cd+ad}\)
(Vì \(a^2+c^2\ge2ac\Leftrightarrow\left(a-c\right)^2\ge0\)luôn đúng; \(b^2+d^2\ge2bd\Leftrightarrow\left(b-d\right)^2\ge0\)luôn đúng)
\(\Leftrightarrow P\ge\frac{2ac+2bd+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+cd+ad+ac+bd}\)
\(\Leftrightarrow P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)
\(\Leftrightarrow F\ge P\ge2\)
\(\LeftrightarrowĐPCM\)
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}.\frac{b^2}{c^2}}=2\frac{a}{c}\\ \frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\frac{c}{b}\\ \frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)
\(=>2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\right)\)
=> đpcm