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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

\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)

\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)

không cs số 0 đâu 

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Sửa: \(\dfrac{3a^2+10b^2-ab}{7a^2+b^2+5ab}=\dfrac{3b^2k^2+10b^2-b^2k}{7b^2k^2+b^2+5b^2k}=\dfrac{b^2\left(3k^2+10-k\right)}{b^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(1\right)\)

\(\dfrac{3c^2+10d^2-cd}{7c^2+d^2+5cd}=\dfrac{3d^2k^2+10d^2-d^2k}{7d^2k^2+d^2+5d^2k}=\dfrac{d^2\left(3k^2+10-k\right)}{d^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)

\(1,\) 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 \("="\Leftrightarrow x=y=z=1\)

câu hỏi? 

Tìm min

\(P=\dfrac{3ab+2b^2}{-b^2+5ab}=\dfrac{b\left(3a+2b\right)}{-b\left(b-5a\right)}=\dfrac{-\left(3a+2b\right)}{b-5a}\)

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 : \(A\le\dfrac{1}{4}\)