Câu 1

\(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\\ \Leftrightarrow N=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{4\left(a+b\right)^2}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

Dấu \("="\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 2:

\(P=a+\dfrac{1}{a}+2b+\dfrac{8}{b}+3c+\dfrac{27}{c}+4\left(a+b+c\right)\\ P\ge2\sqrt{1}+2\sqrt{16}+2\sqrt{81}+4\cdot6=2+8+18+4=32\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\end{matrix}\right.\)

Câu 3: Cho a,b,c là các số thuộc đoạn [ -1;2 ] thõa mãn \(a^2+b^2+c^2=6.\) CMR : \(a+b+c>0\) - Hoc24

\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu \("="\Leftrightarrow x=y\)

\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)

\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)

Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)

Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z=1\)

ĐK : a;b;c > 0 

Ta có : \(ab+bc+ac=1\) \(\Leftrightarrow c\left(a+b\right)=1-ab\Leftrightarrow c=\dfrac{1-ab}{a+b}\)

Khi đó :  \(c^2+1=\left(\dfrac{1-ab}{a+b}\right)^2+1\)  \(=\dfrac{\left(ab\right)^2+1+a^2+b^2}{\left(a+b\right)^2}=\dfrac{\left(a^2+1\right)\left(b^2+1\right)}{\left(a+b\right)^2}\)

\(\Rightarrow\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)^2}{\left(a^2+1\right)\left(b^2+1\right)}\) 

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

Suy ra : \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)\left(ab+1-a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(a+b\right)\left(1-a\right)\left(1-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)

AD BĐT Cauchy ta được :  \(\left(a+b\right)\left[\left(1-a\right)\left(1-b\right)\right]\le\dfrac{\left[a+b+\left(1-a\right)\left(1-b\right)\right]^2}{4}=\dfrac{\left(1+ab\right)^2}{4}\)

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

Suy ra : \(A\le\dfrac{1}{4}\)

Đầu tiên ta chứng minh bđt:\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

Áp dụng \(\Rightarrow P=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\ge\dfrac{4}{a^2+b^2+2ab}=\dfrac{4}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}=\dfrac{1}{4}\)

\(\Rightarrow MINP=\dfrac{1}{4}\Leftrightarrow a=b=2\)

Với \(ab+bc+ca=1\) và a,b,c>0 ta có:

\(\left\{{}\begin{matrix}\sqrt{a^2+1}=\sqrt{\left(a+b\right)\left(c+a\right)}\\\sqrt{b^2+1}=\sqrt{\left(b+c\right)\left(a+b\right)}\\\sqrt{c^2+1}=\sqrt{\left(c+a\right)\left(b+c\right)}\end{matrix}\right.\). Do đó:

\(\dfrac{\sqrt{a^2+1}.\sqrt{b^2+1}}{\sqrt{c^2+1}}=a+b\)

Tương tự: \(\dfrac{\sqrt{b^2+1}.\sqrt{c^2+1}}{\sqrt{a^2+1}}=b+c\) ; \(\dfrac{\sqrt{c^2+1}.\sqrt{a^2+1}}{\sqrt{b^2+1}}=c+a\)

\(\Rightarrow P=2\left(a+b+c\right)\)

\(\Rightarrow P^2=4\left(a+b+c\right)^2\ge4.3\left(ab+bc+ca\right)=4.3.1=12\)

\(\Rightarrow P\ge2\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\)

Vậy \(MinP=2\sqrt{3}\)

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

\(M\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{ab+bc+ca}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow M\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}=9+\dfrac{7.3}{\left(a+b+c\right)^2}=9+21=30\)

\(Min_M=30\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo

\(m\text{≥}\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)

\(=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

≥ \(\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)\(+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

CM BĐT: \(a^2+b^2+c^2\text{≥}ab+bc+ca\)

⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\text{≥}0\) (luôn đúng)

⇒ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ca\right)\)

⇒ \(\dfrac{\left(a+b+c\right)^2}{3}\text{≥}ab+bc+ca\)

⇒ \(m\text{≥}\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=9+21=30\) 

(vì a+b+c=1)

Vậy...

\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\\ =\dfrac{2}{a^2+b^2}+\dfrac{2}{2ab}+\dfrac{34}{ab}+\dfrac{17ab}{8}-\dfrac{ab}{8}\\ =2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+17\left(\dfrac{2}{ab}+\dfrac{ab}{8}\right)-\dfrac{ab}{8}\\ \overset{AM-GM}{\ge}2\cdot\dfrac{1}{a^2+b^2+2ab}+17\sqrt{\dfrac{2}{ab}\cdot\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{2}{\left(a+b\right)^2}+\dfrac{17}{2}-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{2}{4^2}+\dfrac{17}{2}-\dfrac{4^2}{32}=\dfrac{65}{8}\)

Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}\dfrac{2}{ab}=\dfrac{ab}{8}\\a^2+b^2=2ab\\a=b\\a+b=4\end{matrix}\right.\Leftrightarrow a=b=2\)

Vậy \(A_{Min}=\dfrac{65}{8}\) khi \(a=b=2\)

\(\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+17\cdot2\sqrt{\dfrac{2}{ab}+\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{8}{\left(a+b\right)^2}+17-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{8}{4^2}+17-\dfrac{4^2}{32}=17\)

Vậy \(A_{Min}=17\) khi \(a=b=c=2\)

\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{ab}+4ab=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}+8ab-4ab\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{1}{2}.8}-\dfrac{4.\left(a+b\right)^2}{4}=\dfrac{4}{\left(a+b\right)^2}+4-\left(a+b\right)^2\ge4+4-1=7\Rightarrow minA=7\Leftrightarrow a=b=\dfrac{1}{2}\)