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\(R=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{1}=9\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
Vậy GTNN của \(R\) là \(9\) khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt ~
\(A=\dfrac{a}{b+1}+\dfrac{b}{a+1}+\dfrac{1}{a+b}\)
\(\ge\dfrac{a}{a+2b}+\dfrac{b}{2a+b}+\dfrac{1}{a+b}\)
\(=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{2ab+b^2}+\dfrac{1}{a+b}\)
\(\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+2ab}+\dfrac{1}{a+b}\)
\(\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+\dfrac{\left(a+b\right)^2}{2}}+\dfrac{1}{a+b}\)
\(=\dfrac{\left(a+b\right)^2}{\dfrac{3}{2}\left(a+b\right)^2}+\dfrac{1}{a+b}=\dfrac{2}{3}+\dfrac{1}{a+b}\ge\dfrac{2}{3}+1=\dfrac{5}{3}\)
\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)
\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)
\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)
Nhân vế:
\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)
\(\Rightarrow abc\ge8.2017.2018\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2.1;2.2017;2.2018\right)=...\)
Cho 2 số thực dương a,b thỏa mãn \(a+b\le1\) . Tìm GTNN của
\(A=\dfrac{1}{1+a^2+b^2}+\dfrac{1}{2ab}\)
Lời giải:
Áp dụng BĐT AM-GM:
$1\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}$
Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{1}{1+a^2+b^2}+\frac{1}{6ab}+\frac{1}{3ab}\geq \frac{4}{1+a^2+b^2+6ab}+\frac{1}{3ab}\)
\(=\frac{4}{1+(a+b)^2+4ab}+\frac{1}{3ab}\geq \frac{4}{1+1+4.\frac{1}{4}}+\frac{1}{3.\frac{1}{4}}=\frac{8}{3}\)
Vậy $A_{\min}=\frac{8}{3}$ khi $a=b=\frac{1}{2}$
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
Cách khác:
Ta chứng minh: \(\frac{a}{a^3+a+1}\le\frac{1}{2}.\frac{a^{\frac{2}{3}}+1}{a^{\frac{4}{3}}+a^{\frac{2}{3}}+1}\) (1)
Đặt \(a=x^3\Leftrightarrow\frac{x^3}{x^9+x^3+1}\le\frac{1}{2}.\frac{x^2+1}{x^4+x^2+1}\)
Tương đương với $$\frac{(x - 1)^2 (x^9 + 2 x^8 + 4 x^7 + 6 x^6 + 6 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)}{2 (x^2 - x + 1) (x^2 + x + 1) (x^9 + x^3 + 1)} \geq 0$$
Vậy (1) đúng. Thiết lập $3$ bất đẳng thức tương tự và cộng theo vế thu đượcVasc.
\(\Rightarrow\) $\text{đpcm}$
Đẳng thức xảy ra khi $a=b=c=1$
Lời giải:
Xét hiệu: $a^3+1-a(a+1)=a^2(a-1)-(a-1)=(a+1)(a-1)^2\geq 0$ với mọi $a>0$
$\Rightarrow a^3+1\geq a(a+1)\Rightarrow a^3+a+1\geq a(a+2)$
$\Rightarrow \frac{a}{a^3+a+1}\leq \frac{1}{a+2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế thu được:
$\sum \frac{a}{a^3+a+1}\leq \sum \frac{1}{a+2}(*)$
Do $abc=1$ nên tồn tại $x,y,z>0$ sao cho $(a,b,c)=(\frac{x^2}{yz}, \frac{y^2}{xz}, \frac{z^2}{xy})$
Khi đó, áp dụng BĐT Cauchy-Schwarz:
$\sum \frac{1}{a+2}=\sum \frac{yz}{x^2+2yz}=\frac{1}{2}\sum (1-\frac{x^2}{x^2+2yz})=\frac{3}{2}-\frac{1}{2}.\sum \frac{x^2}{x^2+2yz}\leq \frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{x^2+2yz+y^2+2xz+z^2+2xy}$
$=\frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{(x+y+z)^2}=1(**)$
Từ $(*); (**)$ ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=1$
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Đặt \(x=2a\)và \(y=2b\)suy ra \(\hept{\begin{cases}x>0\\y>0\\x+y\le2\end{cases}}\)
Suy ra : \(A=\frac{x}{y+2}+\frac{y}{x+2}+\frac{2}{x+y}\)
\(\Rightarrow A=\frac{x^2}{xy+2x}+\frac{y^2}{xy+2y}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(xy+x+y\right)}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(\frac{\left(x+y\right)^2}{4}+\left(x+y\right)\right)}+\frac{2}{x+y}\)
Đặt \(t=x+y\)( \(0< t\le2\))
Suy ra :
\(\Rightarrow A\ge\frac{t^2}{\frac{t^2}{2}+2t}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{4}{3}.\frac{1}{t}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{2t}{t+4}.\frac{4}{3}.\frac{1}{t}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3\left(t+4\right)}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3.\left(2+4\right)}}+\frac{2}{3}.\frac{1}{2}=\frac{5}{3}\)
"=" xảy ra khi \(x=y=\frac{1}{2}\)