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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 \(x=y=\dfrac{1}{2}\)
1.
\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)
\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)
\(=\left(x^3-x^2+3x\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)
Hay đa thức trên có thể phân tích thành:
\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)
Dựa vào đó em tự tách cho phù hợp
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
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)
\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)
<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)
\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)
áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)
\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)
<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)
dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)
\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)
Từ đó:
\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)
Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)
\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)
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 \(A_{min}=3\Leftrightarrow a=b=c=2\)
Đặt \(\left\{{}\begin{matrix}a+c=x>0\\b+c=y>0\end{matrix}\right.\) \(\Rightarrow xy=1\)
\(A=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{x^2+y^2}{x^2y^2}\)
\(=\dfrac{1}{\left(x-y\right)^2}+x^2+y^2-2xy+2xy\)
\(=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2}}+2=4\)
1.
- Với \(a+b\ge4\Rightarrow A\le0\)
- Với \(a+b< 4\Rightarrow4-a-b>0\)
\(\Rightarrow A=\dfrac{a}{2}.\dfrac{a}{2}.b.\left(4-a-b\right)\)
\(\Rightarrow A\le\dfrac{1}{64}\left(\dfrac{a}{2}+\dfrac{a}{2}+b+4-a-b\right)^4=4\)
\(A_{max}=4\) khi \(\left(a;b\right)=\left(2;1\right)\)
2.
\(P=a+\dfrac{1}{2}.a.2b\left(1+2c\right)\le a+\dfrac{a}{8}\left(2b+1+2c\right)^2\)
\(P\le a+\dfrac{a}{8}\left(7-2a\right)^2=\dfrac{1}{8}\left(4a^3-28a^2+57a-36\right)+\dfrac{9}{2}\)
\(P\le\dfrac{1}{8}\left(a-4\right)\left(2a-3\right)^2+\dfrac{9}{2}\le\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};1;\dfrac{1}{2}\right)\)
Câu 3 bạn xem lại đề, mình có thể chắc chắn với bạn là đề sai
Ví dụ bạn cho \(x=98,y=100\) thì vế trái chỉ lớn hơn 8 một chút
Đề đúng phải là: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)