Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=a+b+c+\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\dfrac{1}{2}\left(a+b+c\right)^2+\left(a+b+c\right)-\dfrac{3}{2}\)
\(A=\dfrac{1}{2}\left(a+b+c+1\right)^2-2\ge-2\)
\(A_{min}=-2\) khi \(a+b+c=-1\) (có vô số bộ a;b;c thỏa mãn điều này)
Với mọi a;b;c ta luôn có:
\(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ca\right)\)
\(\Leftrightarrow12\ge2A\)
\(\Rightarrow A\le6\)
\(A_{max}=6\) khi \(a=b=c=1\)
\(K=\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\left(a,b,c>0\right)\).
Ta có:
\(\frac{a^2}{c\left(a^2+c^2\right)}=\frac{\left(a^2+c^2\right)-c^2}{c\left(a^2+c^2\right)}=\frac{a^2+c^2}{c\left(a^2+c^2\right)}-\frac{c^2}{c\left(a^2+c^2\right)}\)\(=\frac{1}{c}-\frac{c^2}{c\left(a^2+c^2\right)}\).
Vì \(a,c>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(a^2+c^2\ge2ac\).
\(\Leftrightarrow c\left(a^2+c^2\right)\ge2ac^2\).
\(\Rightarrow\frac{1}{c\left(a^2+c^2\right)}\le\frac{1}{2ac^2}\)
\(\Leftrightarrow\frac{c^2}{c\left(a^2+c^2\right)}\le\frac{c^2}{2ac^2}=\frac{1}{2a}\).
\(\Leftrightarrow-\frac{c^2}{c\left(a^2+c^2\right)}\ge-\frac{1}{2a}\).
\(\Leftrightarrow\frac{1}{c}-\frac{c^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\)
\(\Leftrightarrow\frac{a^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\left(1\right)\)
Dấu bằng xảy ra \(\Leftrightarrow a=c>0\) .
Chứng minh tương tự, ta được:
\(\frac{b^2}{a\left(a^2+b^2\right)}\ge\frac{1}{a}-\frac{1}{2b}\left(a,b>0\right)\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow a=b>0\)
Chứng minh tương tự, ta dược:
\(\frac{c^2}{b\left(b^2+c^2\right)}\ge\frac{1}{b}-\frac{1}{2c}\left(b,c>0\right)\left(3\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\ge\)\(\frac{1}{c}-\frac{1}{2a}+\frac{1}{a}-\frac{1}{2b}+\frac{1}{b}-\frac{1}{2c}\).
\(\Leftrightarrow K\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).
\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).
\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{ab+bc+ca}{abc}\right)\).
Mà \(ab+bc+ca=3abc\)(theo đề bài).
Do đó \(K\ge\frac{1}{2}.\frac{3abc}{abc}\).
\(\Leftrightarrow K\ge\frac{3abc}{2abc}\).
\(\Leftrightarrow K\ge\frac{3}{2}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=3abc\end{cases}}\Leftrightarrow a=b=c=1\).
Vậy \(minK=\frac{3}{2}\Leftrightarrow a=b=c=1\).
Áp dụng BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Ta có : \(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{ab}{4\left(a+c\right)}\)
\(+\frac{ab}{4\left(b+c\right)}\)
Thiết lập tương tự và thu lại ta có :
\(P\)\(\le\left[\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}+\frac{bc}{4\left(a+b\right)}+\frac{bc}{4\left(a+c\right)}+\frac{ac}{4\left(a+b\right)}+\frac{ac}{4\left(b+c\right)}\right]\)
\(\Leftrightarrow P\le\frac{ab+bc}{4\left(a+c\right)}+\frac{bc+ac}{4\left(a+b\right)}+\frac{ab+ac}{4\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{b\left(a+c\right)}{4\left(a+c\right)}+\frac{c\left(a+b\right)}{4\left(a+b\right)}+\frac{a\left(b+c\right)}{4\left(b+c\right)}=\frac{a+b+c}{4}=\frac{1}{4}\)
Vậy \(P_{max}=\frac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Áp dụng BĐT : \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Ta có :
\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}\)
Thiết lập tương tự và thu gọn lại ta có :
\(P\le\left[\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}+\frac{bc}{4\left(a+b\right)}+\frac{bc}{4\left(a+c\right)}+\frac{ac}{4\left(a+b\right)}+\frac{ac}{4\left(b+c\right)}\right]\)
\(\Leftrightarrow P\le\frac{ab+bc}{4\left(a+c\right)}+\frac{bc+ac}{4\left(a+b\right)}+\frac{ab+ac}{4\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{b\left(a+c\right)}{4\left(a+c\right)}+\frac{c\left(a+b\right)}{4\left(a+b\right)}+\frac{a\left(b+c\right)}{4\left(b+c\right)}=\frac{1}{4}\)
Vậy \(P_{max}=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
Áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Ta có : \(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}\)
Thiết lập tương tự và thu lại ta có
\(P\le\) \(\left[\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}+\frac{bc}{4\left(a+b\right)}+\frac{bc}{4\left(a+c\right)}+\frac{ac}{4\left(a+b\right)}+\frac{ac}{4\left(b+c\right)}\right]\)
\(\Leftrightarrow P\le\frac{ab+bc}{4\left(a+c\right)}+\frac{bc+ac}{4\left(a+b\right)}+\frac{ab+ac}{4\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{b\left(a+c\right)}{4\left(a+c\right)}+\frac{c\left(a+b\right)}{4\left(a+b\right)}+\frac{a\left(b+c\right)}{4\left(b+c\right)}=\frac{a+b+c}{4}=\frac{1}{4}\)
Vậy \(P_{max}=\frac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
Ta có \(ab+bc+ac\le a^2+b^2+c^2=3\)
\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)
=> \(MaxS=6\)xảy ra khi a=b=c=1
\(2S=2\left(a+b+c\right)+2ab+2bc+2ac+a^2+b^2+c^2-3\)
=> \(2S=2\left(a+b+c\right)+\left(a+b+c\right)^2-3\)
=> \(2S=\left(a+b+c+1\right)^2-4\ge-4\)
=> \(S\ge-2\)
\(MinS=-2\)xảy ra khi a+b+c=-1