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Đặt \(A=\left(\frac{a}{a^2b^2+a^2+1}\right)^2+\left(\frac{b}{b^2c^2+b^2+1}\right)^2+\left(\frac{c}{c^2a^2+c^2+1}\right)^2\)
Cần cm : \(B=\frac{1}{a^2b^2+a^2+1}+\frac{1}{b^2c^2+b^2+1}+\frac{1}{a^2c^2+c^2+1}=1\)
\(B=\frac{a^2b^2c^2}{a^2b^2+a^2+a^2b^2c^2}+\frac{1}{b^2c^2+b^2+1}+\frac{a^2b^2c^2}{a^2c^2+a^2b^2c^3+a^2b^2c^2}\) (Do \(abc=1\))
\(=\frac{b^2c^2}{b^2c^2+b^2+1}+\frac{1}{b^2c^2+b^2+1}+\frac{b^2}{b^2c^2+b^2+1}=\frac{b^2c^2+b^2+1}{b^2c^2+b^2+1}=1\)(đúng)
Ta có : \(A=\frac{\frac{1}{\left(a^2b^2+a^2+1\right)^2}}{a^2}+\frac{\frac{1}{\left(b^2c^2+b^2+1\right)^2}}{b^2}+\frac{\frac{1}{\left(c^2a^2+c^2+1\right)^2}}{c^2}\)
\(\ge\frac{\left(\frac{1}{a^2b^2+a^2+1}+\frac{1}{b^2c^2+b^2+1}+\frac{1}{a^2c^2+c^2+1}\right)^2}{a^2+b^2+c^2}=\frac{B^2}{a^2+b^2+c^2}=\frac{1}{a^2+b^2+c^2}\)(đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
phân thức thức thứ 3 dòng thứ 3 ở mẫu là \(a^2c^2+a^2b^2c^4+a^2b^2c^2\)chứ bạn nhỉ????
\(abc=1\Rightarrow\left(abc\right)^2=a^2b^2c^2=1\Rightarrow a^2=\frac{1}{b^2c^2}\Rightarrow\frac{1}{a^3\left(b+c\right)}=\frac{b^2c^2}{a\left(b+c\right)}=\frac{\left(bc\right)^2}{ab+ac}\)
Chứng minh tương tự ta có: \(\frac{1}{b^3\left(c+a\right)}=\frac{\left(ca\right)^2}{bc+ba};\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{ca+cb}\)
=> \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel: \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{bc+ca+ab+ca+ab+bc}=\frac{ab+bc+ca}{2}\)
Tiếp tục áp dụng bđt Cauchy với 3 số dương ta được: \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3\sqrt[3]{1}}{2}=\frac{3}{2}\)
=> \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{ab+bc+ca}{2}\ge\frac{3}{2}\)
ta có:\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
=\(\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(a+c\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
>= \(\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\)(BĐT Svaxo)=\(\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
>= \(\frac{3\sqrt[3]{a^2b^2c^2}}{2}\left(BĐTAM-GM\right)=\frac{3}{2}\)(đpcm)
dấu = khi a=b=c=1
\(2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2\left(a^2+b^2+c^2\right)+4\frac{ab+bc+ca}{abc}.\)
\(=2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)(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 \(a=b=c=1\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{abc}\Leftrightarrow ab+bc+ac=1\)
\(A=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow1=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).abc\Leftrightarrow1=bc+ac+ab\)
\(A=\left(bc+ac+ab+a^2\right)\left(bc+ac+ab+b^2\right)\left(bc+ac+ab+c^2\right)\)
\(A=\left[c\left(a+b\right)+a\left(a+b\right)\right]\left[c\left(a+b\right)+b\left(a+b\right)\right]\left[c\left(c+b\right)+a\left(c+b\right)\right]\)
\(A=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
Ta có:
\(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\frac{1}{a^2}}{\frac{b+c}{bc}}\)
Biến đổi tương tự cho 2 BĐT còn lại ta có:
\(\frac{1}{b^3\left(a+c\right)}=\frac{\frac{1}{b^2}}{\frac{a+c}{ac}};\frac{1}{c^3\left(a+b\right)}=\frac{\frac{1}{c^2}}{\frac{a+b}{ab}}\)
Cộng theo vế 3 BĐT trên ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{2}{a}+\frac{2}{b}+\frac{2}{c}}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{1}{2}3\sqrt[3]{\left(abc\right)^2}=\frac{3}{2}\)
Từ giả thiết của bài toán, ta biến đổi như sau:
\(a^2+b^2+c^2+\left(a+b+c\right)^2\le4\)
\(\Leftrightarrow a^2+b^2+c^2+ab+ac+bc\le2\)
Bất đẳng thức cần chứng minh tương đương với
\(A=\frac{ab+1}{\left(a+b\right)^2}+\frac{bc+1}{\left(b+c\right)^2}+\frac{ac+1}{\left(a+c\right)^2}\ge3\)
\(\Leftrightarrow\frac{2ab+2}{\left(a+b\right)^2}+\frac{2bc+2}{\left(b+c\right)^2}+\frac{2ac+2}{\left(a+c\right)^2}\ge6\)
Áp dụng giả thiết ta được
\(\frac{2ab+2}{\left(a+b\right)^2}+\frac{2ab+2}{\left(b+c\right)^2}+\frac{2ac+2}{\left(a+c\right)^2}\ge\text{∑}\frac{2ab+a^2+b^2+c^2+ab+bc+ac}{\left(a+b\right)^2}\)
\(=1+\frac{\left(c+a\right)\left(c+b\right)}{\left(a+b\right)^2}+1+\frac{\left(b+a\right)\left(c+b\right)}{\left(a+c^2\right)}+1+\frac{\left(c+a\right)\left(a+b\right)}{\left(c+b\right)^2}\)
\(=3+\frac{\left(c+a\right)\left(c+b\right)}{\left(a+b\right)^2}+\frac{\left(b+a\right)\left(c+b\right)}{\left(a+c\right)^2}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c+b\right)^2}\ge\)
\(3+\sqrt[3]{\frac{\left(c+a\right)\left(c+b\right)\left(b+a\right)\left(c+b\right)\left(c+a\right)\left(a+b\right)}{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}}=3+3=6\)
Vậy bài toán đã được chứng minh. Đẳng thức xảy ra khi và chỉ khi a=b=c=13√.■
sửa đề là chứng minh nó <=1 nha !
ta có \(\frac{2}{\left(a+1\right)^2+b^2+1}=\frac{2}{a^2+b^2+2a+2}\)
mà \(a^2+b^2\ge2ab\Rightarrow\frac{2}{a^2+b^2+2a+2}\le\frac{2}{2ab+2a+2}=\frac{1}{ab+a+1}\)
tương rự, ta có \(...\le\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
mà từ abc=1, ta có thể chứng minh \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=1\)
=>...<=1(ĐPCM)
dấu = xảy ra <=>a=b=c=1
^_^