Ta có: \(\frac{a}{1+b^2}=\frac{a+ab^2-ab^2}{1+b^2}=\frac{a\left(1+b^2\right)}{1+b^2}-\frac{ab^2}{1+b^2}\)

                                                               \(=a-\frac{ab^2}{1+b^2}\)

Áp dụng bđt Cô-si ta có: \(1+b^2\ge2\sqrt{b^2}=2b\)

\(\Rightarrow\frac{ab^2}{1+b^2}\le\frac{ab^2}{2b}=\frac{ab}{2}\)

\(\Rightarrow a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\)

\(\Rightarrow\frac{a}{1+b^2}\ge a-\frac{ab}{2}\)

C/m tương tự \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\)

                     \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng từng vế của 3 bđt trên lại ta đc

\(VT\ge a+b+c-\frac{ab+bc+ca}{2}=3-\frac{ab+bc+ca}{2}\)

Ta có bđt: \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)(1) với x , y , z dương 

Thật vậy \(\left(1\right)\Leftrightarrow\left(x+y+z\right)^2\ge3xy+3yz+3zx\)

                      \(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3xy+3yz+3zx\)

                      \(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx\ge0\)

                    \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

                    \(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(Luôn đúng)

Áp dụng bđt (1) ta đc \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

Khi đó: \(VT\ge3-\frac{3}{2}=\frac{3}{2}\)

Dấu "=" <=> a = b = c = 1

Vậy .............

\(P=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(P\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)

\(P\ge\frac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=\frac{30}{\left(a+b+c\right)^2}=30\)

\(P_{min}=30\) khi \(a=b=c=\frac{1}{3}\)

\(\left(1+a^3\right)\left(1+b^3\right)\left(1+b^3\right)\ge\left(1+ab^2\right)^3\)

\(\Leftrightarrow\)\(\frac{1+a^3}{1+ab^2}\ge\frac{\left(1+ab^2\right)^2}{\left(1+b^3\right)^2}\)

\(\Rightarrow\)\(3P\ge\Sigma\frac{\left(1+ab^2\right)^2}{\left(1+b^3\right)^2}+2\Sigma\frac{1+a^3}{1+ab^2}\ge9\sqrt[9]{\frac{\Pi\left(1+ab^2\right)^2}{\Pi\left(1+a^3\right)^2}\left(\frac{\Pi\left(1+a^3\right)}{\Pi\left(1+ab^2\right)}\right)^2}=9\)

\(\Rightarrow\)\(P\ge3\)

dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)

2 ) Ta có : \(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)

\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)

Do a ; b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\frac{a+b}{3}-1\le0\)

\(\Leftrightarrow a+b\le3\)

\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+\frac{8}{a}+\frac{2}{b}+2b-\left(a+b\right)\ge8+4-3=9\)

( áp dụng BĐT Cauchy cho a ; b dương )

Dấu " = " xảy ra \(\Leftrightarrow a=2;b=1\)

Tìm min cho K, tìm max có lẽ Bunhia là ra thôi:

Đặt \(\left\{{}\begin{matrix}\sqrt{3a+1}=x\\\sqrt{3b+1}=y\\\sqrt{3x+1}=z\end{matrix}\right.\) \(\Rightarrow1\le x;y;z\le\sqrt{10}\)

\(x^2+y^2+z^2=3\left(a+b+c\right)+3=12\)

Bài toán trở thành cho \(x^2+y^2+z^2=12\), tìm min \(P=x+y+z\)

Ta có: \(\left(x-1\right)\left(x-\sqrt{10}\right)\le0\Rightarrow x^2-\left(\sqrt{10}+1\right)x+\sqrt{10}\le0\)

\(\left(y-1\right)\left(y-\sqrt{10}\right)=y^2-\left(\sqrt{10}+1\right)y+\sqrt{10}\le0\)

\(\left(z-1\right)\left(z-\sqrt{10}\right)=z^2-\left(\sqrt{10}+1\right)z+\sqrt{10}\le0\)

Cộng vế với vế:

\(x^2+y^2+z^2-\left(\sqrt{10}+1\right)\left(x+y+z\right)+3\sqrt{10}\le0\)

\(\Rightarrow x+y+z\ge\frac{x^2+y^2+z^2+3\sqrt{10}}{\sqrt{10}+1}=\frac{12+3\sqrt{10}}{\sqrt{10}+1}=2+\sqrt{10}\)

\(\Rightarrow P_{min}=2+\sqrt{10}\) khi \(\left(x;y;z\right)=\left(1;1;\sqrt{10}\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(3;0;0\right)\) và các hoán vị

haha

\(GT\Rightarrow\)\(\frac{1}{a+2}+\frac{3}{b+4}\leq1-\frac{2}{c+3}\)

Áp dụng BĐT AM-GM ta có:

\(1-\frac{2}{c+3}\geq\frac{1}{a+2}+\frac{3}{b+4}\geq2\sqrt{\frac{3}{(a+2)(b+4)}}\)

Tương tự ta có:

\(1-\frac{1}{a+2}\geq\frac{3}{b+4}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(b+4)}}\)

\(1-\frac{3}{b+4}\geq\frac{1}{a+2}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(a+2)}}\)

Nhân theo vế ta được: \((1-\frac{2}{c+3})(1-\frac{1}{a+2})(1-\frac{3}{b+4})\geq \frac{48}{(a+2)(b+4)(c+3)}\)

\(\Leftrightarrow (\frac{c+1}{c+3})(\frac{a+1}{a+2})(\frac{b+1}{b+4})\geq\frac{48}{(a+2)(b+4)(c+3)}\)

\(\Leftrightarrow(a+1)(b+1)(c+1)\geq48\)

Dấu "=" xảy ra khi \(a=1;c=3;b=5\)

\(Gt\Leftrightarrow 1-\frac{1}{a+2}+1-\frac{3}{b+4}+\frac{c+1}{c+3}\geq 2\\\Leftrightarrow \frac{a+1}{a+2}+\frac{b+1}{b+4}+\frac{c+1}{c+3}\geq 2\)

Đặt \((a+1;b+1;c+1)\rightarrow (x;y;z)\), vậy cần tìm GTNN của \(Q=xyz\)

Ta có: \(\frac{x}{x+1}+\frac{y}{y+3}+\frac{z}{z+2}\geq 2\)

Áp dụng BĐT AM-GM ta có:

\(\frac{x}{x+1}\geq 1-\frac{y}{y+3}+1-\frac{z}{z+2}=\frac{3}{y+3}+\frac{2}{z+2}\geq 2\sqrt{\frac{6}{(y+3)(z+2)}}\)

\(\frac{y}{y+3}\geq 1-\frac{x}{x+1}+1-\frac{z}{z+2}=\frac{1}{x+1}+\frac{2}{z+2}\geq 2\sqrt{\frac{2}{(x+1)(z+2)}}\)

\(\frac{z}{z+2}\geq 1-\frac{x}{x+1}+1-\frac{y}{y+3}= \frac{1}{x+1}+\frac{3}{y+3}\geq 2\sqrt{\frac{3}{(x+1)(y+3)}}\)

Nhân theo vế ta có:\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{48}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\Leftrightarrow Q\ge48\)

Dấu "=" xảy ra khi \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{x+1}=\frac{3}{y+3}=\frac{2}{z+2} & & \\ \frac{1}{a+2}+\frac{3}{b+4}=\frac{c+1}{c+3} & & \end{matrix}\right.\)\(\Leftrightarrow a=1;b=5;c=3\)

mk còn hơi nhiều cách giải cần tham khảo thì pm

Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)

\(\ge\left(1+3a\right)\left(1-\frac{b^2}{2b}\right)=\left(1+3a\right)\left(1-\frac{b}{2}\right)\)

\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)

Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)

Đẳng thức xảy ra khi a = b = c = 1

\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)

Ta tách VT=A+B và xét

\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\text{∑}\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\text{∑}\left(3a-\frac{3ab}{2}\right)\)

\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\text{∑}\left(1-\frac{b^2}{1+b^2}\right)\ge\text{∑}\left(1-\frac{b}{2}\right)\)

\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\text{∑}ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)

(Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))

Dấu = khi a=b=c=1

2 + 2 =22

5555555555555 = 5 x 5 x ........