Biểu thức này có vẻ chỉ tìm được min chứ ko tìm được max:

Min:

\(P^2=a+b+c+a^3b^3+b^3c^3+c^3a^3+2\sqrt{\left(a+b^3c^3\right)\left(b+c^3a^3\right)}+2\sqrt{\left(a+b^3c^3\right)\left(c+a^3b^3\right)}+2\sqrt{\left(b+c^3a^3\right)\left(c+a^3b^3\right)}\)

\(P^2\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\ge a+b+c=2\)

\(\Rightarrow P\ge\sqrt{2}\)

\(P_{min}=\sqrt{2}\) khi \(\left(a;b;c\right)=\left(0;0;2\right)\) và các hoán vị

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

1,\(T=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=20\left(a^2-ab+b^2\right)=\)

\(=10\left(a^2-2ab+b^2\right)+10\left(a^2+b^2\right)\)

\(\ge10\left(a-b\right)^2+5.\left(a+b\right)^2\ge0+5.20^2=2000\)

2,a,\(\sqrt{a}+\sqrt{b-1}+\sqrt{c-2}=\frac{1}{2}\left(a+b+c\right)\)

\(\Leftrightarrow a-2\sqrt{a}+b-2\sqrt{b-1}+c-2\sqrt{c-2}=0\)

\(\Leftrightarrow a-2\sqrt{a}+1+b-1-2\sqrt{b-1}+1+c-2+2\sqrt{c-2}+1=0\)

\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-1}-1\right)^2+\left(\sqrt{c-2}-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)

b,sai đề

Xét \(\frac{a+b}{2}\ge\sqrt{ab}\Rightarrow10\ge\sqrt{ab}\Leftrightarrow100\ge ab\)

\(T=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=20\left(a^2-ab+b^2\right)=20\left[a^2+2ab+b^2-3ab\right]=20\left(20\right)^2-6ab\)

\(T\ge20.20^2-6.100=7400\)

b. \(1=\left(a+2b\right)^2\ge4.a.2b=8ab\)

\(\Rightarrow ab\le\frac{1}{8}\)

Dấu = xảy ra khi \(a=\frac{1}{2}\);\(b=\frac{1}{8}\)

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 

Ta có bất đẳng thức phụ sau (bđt Mincopski)

\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\left(x;y;z;t\inℝ\right)\)

Thật vậy :

 \(bđt\Leftrightarrow x^2+y^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+z^2+t^2\ge x^2+2xz+z^2+y^2+2yt+t^2\)

\(\Leftrightarrow\sqrt{x^2z^2+x^2t^2+y^2z^2+y^2t^2}\ge xz+yt\)

*Nếu xz + yt < 0 thì bđt hiển nhiên đúng

*Nếu xz + yt > 0 thì bđt trở thành 

\(x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2z^2+2xyzt+y^2t^2\)

\(\Leftrightarrow x^2t^2-2xyzt+y^2z^2\ge0\)

\(\Leftrightarrow\left(xt-yz\right)^2\ge0\)(ĐÚng)

Vậy bđt được chứng minh

Áp dụng bđt trên 2 lần ta được

\(P\ge\sqrt{\left(5+5\right)^2+\left(a^2+b^2\right)^2}+\sqrt{25+c^4}\)

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

   \(=\sqrt{225+\left(a^2+b^2+c^2\right)^2}\)

Bài toán quay về tìm \(min\left(a^2+b^2+c^2\right)\)biết \(2\left(a+b+c\right)+ab+bc+ca=18\)

Ta có bđt phụ sau \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)(Tự chứng minh bằng biến đổi tương đương nhé)

        \(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)

Đặt \(3\left(a^2+b^2+c^2\right)=t\left(t\ge0\right)\)

\(\Rightarrow a+b+c\le\sqrt{3t}\)

Lại có bđt phụ sau \(ab+bc+ca\le a^2+b^2+c^2=\frac{t}{3}\)

Tóm lại ta thu được 2 bđt sau \(\hept{\begin{cases}a+b+c\le\sqrt{3t}\\ab+bc+ca\le\frac{t}{3}\end{cases}}\)

Ta có \(18=2\left(a+b+c\right)+ab+bc+ca\le2\sqrt{3t}+\frac{t}{3}\)

\(\Leftrightarrow\frac{t}{3}+2\sqrt{3t}-18\ge0\)

\(\Leftrightarrow t+6\sqrt{3t}-54\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{t}\le-9-3\sqrt{3}\left(Loa_.i\cdot do\cdot\sqrt{t}\ge0\right)\\\sqrt{t}\ge9-3\sqrt{3}\left(Tm\right)\end{cases}}\)

Có \(\sqrt{t}\ge9-3\sqrt{3}\)

\(\Leftrightarrow\sqrt{3\left(a^2+b^2+c^2\right)}\ge9-3\sqrt{3}\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge108-54\sqrt{3}\)

\(\Leftrightarrow a^2+b^2+c^2\ge36-18\sqrt{3}\)

Quay trở lại bài toán \(P\ge\sqrt{225+\left(a^2+b^2+c^2\right)^2}\ge\sqrt{225+\left(36-18\sqrt{3}\right)^2}\)

Dấu "=" xảy ra tại a = b = c

P/S: sai đâu thì thôi nha :v a lười ktra lại lắm

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 \(x=y=\dfrac{1}{2}\)

\(b^4+c^4-bc\left(b^2+c^2\right)=\left(b^2+bc+c^2\right)\left(b-c\right)^2\)

\(\Rightarrow b^4+c^4\ge bc\left(b^2+c^2\right)\)

Tương tự\(\Rightarrow\Sigma_{cyc}\frac{a}{a+b^4+c^4}\le\Sigma_{cyc}\frac{a}{a+bc\left(b^2+c^2\right)}=\Sigma_{cyc}\frac{a}{bc\left(a^2+b^2+c^2\right)}=\frac{1}{a^2+b^2+c^2}\Sigma_{cyc}\frac{a}{bc}\)

\(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}=a^2+b^2+c^2\)

\(\Rightarrow\frac{1}{a^2+b^2+c^2}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)=1\)

oke rồi he

@Nub :v

Áp dụng Bunhiacopski ta dễ có:

\(\frac{a}{b^4+c^4+a}=\frac{a\left(1+1+a^3\right)}{\left(b^4+c^4+a\right)\left(1+1+a^3\right)}\le\frac{a^4+2a}{\left(a^2+b^2+c^2\right)^2}\)

Tương tự:

\(\frac{b}{a^4+c^4+b}\le\frac{b^4+2b}{\left(a^2+b^2+c^2\right)^2};\frac{c}{a^4+b^4+c}\le\frac{c^4+2c}{\left(a^2+b^2+c^2\right)^2}\)

Cộng lại:

\(A\le\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\)

Ta đi chứng minh:

\(\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\le1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)

Cái này luôn  đúng theo Cauchy

Đẳng thức xảy ra tại a=b=c=1