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\(VT\leΣ\frac{1}{a^2+b^2+1}\le\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\le\frac{\left(Σa\right)^2}{\left(Σa\right)^2}=1=VP\)
\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\left(a,b,c>0\right)\).
Với \(a,b>0\), ta có:
\(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\).
\(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\).
\(\Leftrightarrow a^4-a^3-a+1\ge0\).
\(\Leftrightarrow a^4-a^3+1\ge a\).
\(\Leftrightarrow a^4-a^3+ab+2\ge ab+a+1\).
\(\Leftrightarrow\sqrt{a^4-a^3+ab+2}\ge\sqrt{ab+a+1}\).
\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\left(1\right)\).
Dấu bằng xảy ra \(\Leftrightarrow a-1=0\Leftrightarrow a=1\).
Chứng minh tương tự (với \(b,c>0\)), ta được:
\(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\left(2\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=1\).
Chứng minh tương tự (với \(a,c>0\)), ta được:
\(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+a+1}}\left(3\right)\)
Dấu bằng xảy ra \(\Leftrightarrow c=1\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\left(4\right)\).
Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki cho 3 số, ta được:
\(\left(1.\frac{1}{\sqrt{ab+a+1}}+1.\frac{1}{\sqrt{bc+b+1}}+1.\frac{1}{\sqrt{ca+c+1}}\right)^2\)\(\le\)\(\left(1^2+1^2+1^2\right)\)\(\left[\frac{1}{\left(\sqrt{ab+a+1}\right)^2}+\frac{1}{\left(\sqrt{bc+b+1}\right)^2}+\frac{1}{\left(\sqrt{ca+c+1}\right)^2}\right]\).
\(\Leftrightarrow\left(\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\right)^2\)\(\le3\left(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)\).
Ta có:
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{c}{abc+ac+c}+\frac{abc}{bc+b+abc}+\frac{1}{ca+c+1}\)(vì \(abc=1\)).
\(=\frac{c}{1+ac+c}+\frac{abc}{b\left(c+1+ac\right)}+\frac{1}{ca+c+1}\)(vì \(abc=1\)).
\(=\frac{c}{1+ac+c}+\frac{ac}{1+ac+c}+\frac{1}{1+ac+c}=1\).
Do đó:
\(\left(\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\right)^2\le3.1=3\).
\(\Leftrightarrow\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\le\sqrt{3}\left(5\right)\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\)\(\sqrt{3}\)(điều phải chứng minh).
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=1\).
Vậy \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\le\sqrt{3}\)với \(a,b,c>0\)và \(abc=1\).
\(+2\)nhé, không phải \(-2\)đâu.
Ta có
\(\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(\Leftrightarrow\frac{2a}{\sqrt{ab+bc+ca+a^2}}+\frac{b}{\sqrt{ab+bc+ca+b^2}}+\frac{c}{\sqrt{ab+bc+ca+c^2}}\)
\(\Leftrightarrow2a.\frac{1}{\sqrt{\left(a+b\right)\left(a+c\right)}}+b.\frac{1}{\sqrt{\left(b+a\right)\left(b+c\right)}}+c.\frac{1}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(\Leftrightarrow2a.\frac{1}{\sqrt{\left(a+b\right)\left(a+c\right)}}+2b.\frac{1}{\sqrt{\left(a+b\right).4.\left(b+c\right)}}+2c.\frac{1}{\sqrt{\left(a+c\right).4.\left(b+c\right)}}\)
\(\le\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{4\left(b+c\right)}+\frac{c}{a+c}+\frac{c}{4\left(b+c\right)}\)
\(=1+1+\frac{1}{4}=\frac{9}{4}\)
Đặt: \(M=\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}=\Sigma_{cyc}\frac{a}{a^2+ab+bc+ca}\)
\(\Rightarrow M.\left(a+b+c\right)=3-\Sigma_{cyc}\frac{bc}{a^2+ab+bc+ca}\)
Đến đây t cần chứng minh:
\(\frac{bc}{a^2+ab+bc+ca}+\frac{ca}{b^2+ab+bc+ca}+\frac{ab}{c^2+ab+bc+ca}\ge\frac{3}{4}\) (*)
Từ điều kiện ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\left(x,y,z>0\right)\)
\(\Rightarrow x+y+z=1\)
(*) \(\Leftrightarrow\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{y^2}{\left(x+y\right)\left(y+z\right)}+\frac{z^2}{\left(y+z\right)\left(z+x\right)}\ge\frac{3}{4}\)
Theo Cô-si: \(\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{9}{16}\left(x+y\right)\left(z+x\right)\ge\frac{3}{2}x\)
Nhứng phần kia tương tự
\(\Rightarrow\Sigma_{cyc}\frac{x^2}{\left(x+y\right)\left(z+x\right)}\ge\frac{3}{2}\left(x+y+z\right)-\frac{9}{16}\left[\left(x+y+z\right)^2+\left(xy+yz+zx\right)\right]\ge\frac{3}{4}\)
Lần trước làm không đúng hy vọng bây giờ gỡ lại được
Đề bài bị nhầm phải ko bạn.
Ta đặt P=\(\dfrac{b^3}{a}+\dfrac{a^3}{c}+\dfrac{c^3}{b}\) .Ta cần chứng minh P\(\ge3\)\(\dfrac{b^3}{a}+ab\ge2b^2;\dfrac{a^3}{c}+ac\ge2a^2;\dfrac{c^3}{b}+bc\ge2c^2\Rightarrow\dfrac{b^3}{a}+\dfrac{a^3}{c}+\dfrac{c^3}{b}\ge2a^2+2b^2+2c^2-ab-ca-bc\ge ab+bc+ca\Rightarrow2\cdot P\ge2ab+2bc+2ca\left(1\right)\) \(\dfrac{b^3}{a}+a+1\ge3b;\dfrac{a^3}{c}+c+1\ge3a;\dfrac{c^3}{b}+b+1\ge3c\Rightarrow\dfrac{b^3}{a}+\dfrac{a^3}{c}+\dfrac{c^3}{b}\ge3a+3b+3c-3-a-b-c=2a+2b+2c-3\left(2\right)\) Cộng từng vế của 2 bđt (1) và (2) ta được:
\(\Rightarrow3\cdot\left(\dfrac{b^3}{a}+\dfrac{a^3}{c}+\dfrac{c^3}{b}\right)\ge2\left(a+b+c+ab+bc+ca\right)-3=12-3=9\Rightarrow3P\ge9\Rightarrow P\ge3\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng bđt AM-GM :
\(\frac{1}{a^2+1}+\frac{a^2+1}{4}\ge2\sqrt{\frac{a^2+1}{\left(a^2+1\right)\cdot4}}=1\)
Tương tự ta có :
\(\frac{1}{b^2+1}+\frac{b^2+1}{4}\ge1\)
\(\frac{1}{c^2+1}+\frac{c^2+1}{4}\ge1\)
Cộng từng vế ta có :
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}+\frac{a^2+b^2+c^2+3}{4}\ge3\)
Áp dụng bđt quen thuộc : \(a^2+b^2+c^2\ge ab+bc+ac=3\)
Khi đó : \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge3-\frac{3+3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
bạn làm sai rồi . Khi \(a^2+b^2+c^2\ge3\) bạn chuyển vế thì nó không cùng dấu với bất đẳng thức
\(\frac{ab}{a^2+b^2}\le\frac{ab}{2ab}=\frac{1}{2}\)
tương tự \(\frac{\Rightarrow ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ac}{a^2+c^2}\le\frac{3}{2}\)
=>Thắng Nguyễn :cm theo cách đó sai
Ta có \(\left(a+b+c\right)\left(ab+bc+ca\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
Mà \(abc\le\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\) (AM-GM)
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)+\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Rightarrow\frac{9}{8}\ge\left(a+b+c\right)\left(ab+bc+ca\right)\ge\sqrt{3\left(ab+bc+ca\right)}.\left(ab+bc+ca\right)\)
\(\Rightarrow3\left(ab+bc+ca\right)^3\le\frac{81}{64}\)
\(\Rightarrow ab+bc+ca\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)
Ta có: \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=1\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)-abc=1\)
Áp dụng BĐT Cô si ta có
\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow abc\le\frac{1}{8}\)
mặt khác: \(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\left(\frac{2a+2b+2c}{3}\right)^3\)
\(\Rightarrow a+b+c\ge\frac{3}{2}\)
\(\Rightarrow ab+bc+ca=\frac{1+abc}{a+b+c}\le\frac{1+\frac{1}{8}}{\frac{3}{2}}=\frac{3}{4}\)