Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

1a

\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)

\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(A_{min}=\frac{161}{16}\)

1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)

\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)

Bài 2 Dùng Cauchy-Schwarz dạng Engel là ra:D

Bài 3:Đừng vội dùng Cauchy-Schwarz dạng Engel ngay kẻo bị phức tạp:v Thay vào đó hãy khai triển nó ra:

\(A=x^2+y^2+2\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{x^2}+\frac{1}{y^2}\)

\(\ge4+2.2+\frac{4}{x^2+y^2}=4+4+1=9\)

Đẳng thức xảy ra khi \(x=y=\sqrt{2}\)

Bài 4: Dùng Cauchy or Bunhiacopxki là ok!

\(Ta có: \frac{{a^5 }}{{b^3 + c^2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }}\mathop \ge \frac{{3a^2 }}{2}\)

\(\Rightarrow \frac{{a^5 }}{{b^3 + c^2 }} \ge \frac{{3a^2 }}{2} - (\frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }})\)

\(Do đó: \frac{{a^5 }}{{b^3 + c^2 }} \ge \frac{{3a^2 }}{2} - \frac{{\sqrt {2a(b^3 + c^2 )} }}{2}\mathop \ge \frac{{3a^2 }}{2} - \frac{{2a + b^3 + c^2 }}{4}\)

\(CMTT \frac{{b^5 }}{{c^3 + a^2 }}\mathop \ge \frac{{3b^2 }}{2} - \frac{{2b + c^3 + a^2 }}{4}\), \(\frac{{c^5}}{{a^3+b^2}}\mathop \ge \frac{{3c^2 }}{2} - \frac{{2c + a^3 + b^2 }}{4}\)

\(M \ge \frac{{3(a^2 + b^2 + c^2 )}}{2} + a^4 + b^4 + c^4 - \frac{{2(a + b + c) + (a^2 + b^2 + c^2 ) + (a^3 + b^3 + c^3 )}}{4}\)

\(M \ge \frac{9}{2} + a^4 + b^4 + c^4 - \frac{{2(a + b + c) + (a^2 + b^2 + c^2 ) + (a^3 + b^3 + c^3 )}}{4}\)

Áp dụng Bunhiacoopski ta có:

\(\sqrt {(a^4+b^4+c^4 )(a^2+b^2+c^2)}=\sqrt {(a^4 +b^4+ c^4 ).3}\ge a^3+b^3+c^3 \)

\(\sqrt {(a^4 + b^4 + c^4 )(1 + 1 + 1)} = \sqrt {(a^2 + b^2 + c^2 ).3} \ge a^2 + b^2 + c^2 \Leftrightarrow a^4 + b^4 + c^4 \ge 3\)

Ta có: \(3 = a^2 + b^2 + c^2 \ge \frac{{(a + b + c)^2 }}{3} \Leftrightarrow a^2 + b^2 + c^2 \ge a + b + c\) 

\(Đặt t=x^4+y^4+z^4 (t \ge 3) cần CM để trở thành S \ge \frac{{4t - 9 - \sqrt {3t} }}{4}\ge 0\)

\(Ta có: S\ge \frac{{4t - 9 - \sqrt {3t} }}{4} = \frac{{3(t - 3) + \sqrt t (\sqrt t - \sqrt 3 )}}{4} \ge 0 \)
\(Do đó: M\geq \frac{9}{2}\)

Phần đầu mình thiếu nha

\(\frac{a^5}{b^3+c^2}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}\ge\frac{3a^2}{2}\)

=> \(\frac{a^5}{b^3+c^2}\ge\frac{3a^2}{2}-\left(\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}\right)\)

Do đó \(\frac{a^5}{b^3+c^2}\ge\frac{3a^2}{2}-\frac{\sqrt{2a\left(b^3+c^2\right)}}{2}\ge\frac{3a^2}{2}-\frac{\left(2a+b^3+b^2\right)}{4}\)

CMTT \(\frac{b^5}{c^3+a^2}\ge\frac{3b^2}{2}-\frac{\left(2b+c^3+a^2\right)}{4},\frac{c^5}{a^3+b^2}\ge\frac{3c^2}{2}-\frac{\left(2c+a^3+b^2\right)}{4}\)

Ta có:

\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a}{2}\ge\dfrac{3a^2}{2}\)

\(\Rightarrow M\ge\dfrac{3}{2}\left(a^2+b^2+c^2\right)+a^4+b^4+c^4-\dfrac{1}{4}\left(a^2+b^2+c^2+a^3+b^3+c^3\right)-\dfrac{1}{2}\left(a+b+c\right)\)

\(\ge\dfrac{5}{4}\left(a^2+b^2+c^2\right)+\dfrac{4}{3}\left(a^3+b^3+c^3\right)-1-\dfrac{1}{4}\left(a^3+b^3+c^3\right)-\dfrac{1}{4}\left(a^2+b^2+c^2\right)-\dfrac{3}{4}\)

\(=\left(a^2+b^2+c^2\right)+\dfrac{13}{12}\left(a^3+b^3+c^3\right)-\dfrac{7}{4}\)

\(=\dfrac{5}{4}+\dfrac{13}{12}\left(a^3+b^3+c^3\right)\)

\(\ge\dfrac{5}{4}+\dfrac{3}{2}.\dfrac{13}{12}\left(a^2+b^2+c^2-1\right)=\dfrac{9}{2}\)

Dấu = xảy ra khi \(a=b=c=1\)      

Kiểm tra lại mẫu số của 3 phân thức

Mẫu số của \(b+1\ne c+2,a+2.\)

Xem lại đề bạn

Tôi học chuyên toán nha

Áp dụng BĐT Cô-si:

\(a^2+3\ge2\sqrt{3a^2}=2\sqrt{3}a\)

Tương tự: \(b^2+3\ge2\sqrt{3}b\) ; \(c^2+3\ge2\sqrt{3}c\)

Cộng vế: \(a^2+b^2+c^2+9\ge2\sqrt{3}\left(a+b+c\right)\)

\(\Rightarrow a+b+c\le\dfrac{a^2+b^2+c^2+9}{2\sqrt{3}}=\dfrac{9+9}{2\sqrt{3}}=3\sqrt{3}\)

\(\Rightarrow-\left(a+b+c\right)\ge-3\sqrt{3}\)

Tiếp tục áp dụng BĐT Cô-si:

\(\dfrac{a^4}{b+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(b+2\right)\ge2\sqrt{\dfrac{9a^4\left(b+2\right)}{\left(b+2\right)\left(2+\sqrt{3}\right)^2}}=\dfrac{6a^2}{2+\sqrt{3}}\) 

Tương tự:

\(\dfrac{b^4}{c+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(c+2\right)\ge\dfrac{6b^2}{2+\sqrt{3}}\)

\(\dfrac{c^4}{a+2}+\dfrac{9}{\left(2+\sqrt{3}\right)}\left(a+2\right)\ge\dfrac{6c^2}{2+\sqrt{3}}\)

Cộng vế:

\(P+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{6}{2+\sqrt{3}}\left(a^2+b^2+c^2\right)=\dfrac{54}{2+\sqrt{3}}\)

\(\Rightarrow P\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}.\left(3\sqrt{3}+6\right)\)

\(\Rightarrow P\ge\dfrac{27}{2+\sqrt{3}}=27\left(2-\sqrt{3}\right)\)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)

Làm được bài này chưa. @@@

Chưa, cậu làm được chưa?

Vừa nãy , em tớ là giảng viên đại học trong Miền Nam ra chơi cũng chào thua.

Dự đoán dấu "=" xảy ra khi \(a=b=c=1\)

Khi đó \(\frac{a^4}{b+2}=\frac{1}{3}\)

Ta cần ghép \(\frac{a^4}{b+2}\)với hạng tử \(k\left(b+2\right)\)thỏa mãn khi Cô-si thì dấu "=" xảy ra khi \(a=b=1\)

Lại có \(b+2=3\)

Đồng thời khi Cô-si dấu "=" xảy ra khi \(\frac{a^4}{b+2}=k\left(b+2\right)\)hay \(\frac{1}{3}=k.3\)\(\Leftrightarrow k=\frac{1}{9}\)

Áp dụng BĐT Cô-si cho 2 số dương \(\frac{a^4}{b+2}\)và \(\frac{b+2}{9}\), ta có:

\(\frac{a^4}{b+2}+\text{​​}\frac{b+2}{9}\ge2\sqrt{\frac{a^4}{b+2}.\frac{b+2}{9}}=\frac{2a^2}{3}\)

Tương tự, ta có \(\frac{b^4}{c+2}+\text{​​}\frac{c+2}{9}\ge2\sqrt{\frac{b^4}{c+2}.\frac{c+2}{9}}=\frac{2b^2}{3}\)và 

\(\frac{c^4}{a+2}+\text{​​}\frac{a+2}{9}\ge2\sqrt{\frac{c^4}{a+2}.\frac{a+2}{9}}=\frac{2c^2}{3}\)

CỘng vế theo vế từng BĐT, ta được \(P+\frac{a+2+b+2+c+2}{9}\ge\frac{2\left(a^2+b^2+c^2\right)}{3}\)

\(\Leftrightarrow P+\frac{\left(a+b+c\right)+6}{9}\ge2\)(vì \(a^2+b^2+c^2=3\)) \(\Leftrightarrow P\ge2-\frac{\left(a+b+c\right)+6}{9}\)(1)

Ta chứng minh BĐT phụ \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)(với \(a,b,c>0\))

Thật vậy, BĐT này \(\Leftrightarrow\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\le3a^2+3b^2+3c^2\)\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)

Vậy BĐT phụ được chứng minh \(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=\sqrt{3.3}=3\)(2)

Từ (1) và (2) \(\Rightarrow P\ge2-\frac{3+6}{9}=1\)\(\Rightarrow min_P=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

t ko bic

Áp dụng BĐT Cô-si:

\(\left(a^2+1\right)+\left(b^2+1\right)+\left(c^2+1\right)\ge2a+2b+2c\)

\(\Rightarrow a+b+c\le\dfrac{a^2+b^2+c^2+3}{2}=3\)

\(\Rightarrow-\left(a+b+c\right)\ge-3\)

Vẫn áp dụng Cô-si:

\(\dfrac{a^4}{b+2}+\dfrac{b+2}{9}\ge2\sqrt{\dfrac{a^4\left(b+2\right)}{\left(b+2\right)9}}=\dfrac{2a^2}{3}\)

Tương tự: \(\dfrac{b^4}{c+2}+\dfrac{c+2}{9}\ge\dfrac{2b^2}{3}\) ; \(\dfrac{c^4}{a+2}+\dfrac{a+2}{9}\ge\dfrac{2c^2}{3}\)

Cộng vế: 

\(P+\dfrac{a+b+c+6}{9}\ge\dfrac{2}{3}\left(a^2+b^2+c^2\right)=2\)

\(\Rightarrow P\ge2-\dfrac{1}{9}\left(a+b+c\right)-\dfrac{6}{9}\ge2-\dfrac{1}{9}.3-\dfrac{6}{9}=1\)

\(P_{min}=1\) khi \(a=b=c=1\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)

\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)

à nhầm, \(a=b=c=\frac{4}{3}\) nhé