Ta có \(a+b+c\le\sqrt{3}\)

\(\Rightarrow\left(a+b+c\right)^2\le3\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\le1\)

Theo hệ quả của bất đẳng thức Cauchy\

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)

\(\Rightarrow1\ge ab+bc+ca\)

\(\Rightarrow\left\{\begin{matrix}1+a^2\ge a^2+ab+bc+ca=\left(a+c\right)\left(a+b\right)\\1+b^2\ge b^2+ab+bc+ca=\left(b+c\right)\left(a+b\right)\\1+c^2\ge c^2+ab+bc+ca=\left(b+c\right)\left(c+a\right)\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\sqrt{1+a^2}\ge\sqrt{\left(a+c\right)\left(a+b\right)}\\\sqrt{1+b^2}\ge\sqrt{\left(b+c\right)\left(a+b\right)}\\\sqrt{1+c^2}\ge\sqrt{\left(b+c\right)\left(c+a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a^2+1}}\le\frac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\frac{b}{\sqrt{1+b^2}}\le\frac{b}{\sqrt{\left(b+c\right)\left(a+b\right)}}\\\frac{c}{\sqrt{1+c^2}}\le\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\end{matrix}\right.\)

\(\Rightarrow VT\le\frac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(a+b\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)

\(\Leftrightarrow VT\le\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\) ( 1 )

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left\{\begin{matrix}2\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}\le\frac{a}{a+c}+\frac{a}{a+b}\\2\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}\le\frac{b}{b+c}+\frac{b}{a+b}\\2\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\le\frac{c}{b+c}+\frac{c}{c+a}\end{matrix}\right.\)

\(\Rightarrow2\left(\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\right)\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}+\frac{a}{a+c}\)

\(\Rightarrow2\left(\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\right)\le1+1+1=3\)

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

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{3}{2}\) ( đpcm )

@phynit

\(VT=a+b+c=\alpha.\frac{a}{\alpha}+\beta.\frac{b}{\beta}+\gamma.\frac{c}{\gamma}\)

Áp dụng phương pháp nhóm ABEL

\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\alpha}+\frac{b}{\beta}\ge2\sqrt{\frac{ab}{\alpha\beta}}\left(1\right)\\\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\left(3\right)\end{matrix}\right.\)

Ta có \(ab\ge\alpha\beta\Rightarrow\frac{ab}{\alpha\beta}\ge1\) \(\Rightarrow2\sqrt{\frac{ab}{\alpha\beta}}\ge2\left(2\right)\)

Ta có \(abc\ge\alpha\beta\gamma\Rightarrow\frac{abc}{\alpha\beta\gamma}\ge1\Rightarrow3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\ge3\left(4\right)\)

Từ ( 1 ) và ( 2 )

\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}\ge2\)

\(\Rightarrow\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)\ge2\left(\beta-\gamma\right)\) ( 5 )

Từ ( 3 ) và ( 4 )

\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\)

\(\Rightarrow\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge3\gamma\) ( 6 )

Theo đề bài ta có \(a\ge\alpha\Rightarrow\frac{a}{\alpha}\ge1\)\(\Rightarrow\left(\alpha-\beta\right)\frac{a}{\alpha}\ge\alpha-\beta\) ( 7 )

Từ ( 5 ) , ( 6 ) , ( 7 ) cộng theo từng vế

\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge2\left(\beta-\gamma\right)+3\gamma+\alpha-\beta\)

\(\Rightarrow VT\ge2\beta-2\gamma+3\gamma+\alpha-\beta\)

\(\Rightarrow VT\ge\alpha+\beta+\gamma\)

\(\Leftrightarrow a+b+c\ge\alpha+\beta+\gamma\) ( đpcm )

a)

Theo đề bài ta có: \(\frac{a+4}{b+10}=\frac{a}{b}\)

\(\Leftrightarrow b\left(a+4\right)=a\left(b+10\right)\)

\(\Leftrightarrow ab+4b=ab+10a\)

\(\Leftrightarrow4b=10a\)

\(\Rightarrow\frac{a}{b}=\frac{4}{10}=\frac{2}{5}\)

Vậy \(\frac{a}{b}=\frac{2}{5}\)

b)

Theo đề bài ta có: \(\frac{a+b}{2b}=\frac{2a}{b}\)

\(\Leftrightarrow b\left(a+b\right)=4ab\)

\(\Leftrightarrow ab+b^2=4ab\)

\(\Leftrightarrow b^2=3ab\)

\(\Leftrightarrow b=3a\)

\(\Rightarrow\frac{a}{b}=\frac{1}{3}\)

Vậy phân số \(\frac{a}{b}=\frac{1}{3}\)

Xét: \(\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\)

\(\Leftrightarrow\frac{2a+b+c}{\left(a+b+c\right)a+bc}+\frac{a+2b+c}{\left(a+b+c\right)b+ca}+\frac{a+b+2c}{\left(a+b+c\right)c+ab}\)

\(\Leftrightarrow\frac{2a+b+c}{a^2+ab+ca+bc}+\frac{a+2b+c}{ab+b^2+bc+ca}+\frac{a+b+2c}{ac+bc+c^2+ab}\)

\(\Leftrightarrow\frac{2a+b+c}{a\left(a+b\right)+c\left(a+b\right)}+\frac{a+2b+c}{b\left(b+a\right)+c\left(b+a\right)}+\frac{a+b+2c}{c\left(a+c\right)+b\left(a+c\right)}\)

\(\Leftrightarrow\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}+\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}+\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{matrix}\right.\)

\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Xét: \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\ge\frac{\left(2+2+2\right)^2}{2a+b+c+a+2b+c+a+b+2c}=\frac{36}{4\left(a+b+c\right)}=\frac{36}{12}=3\)

Mà \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

\(\Rightarrow VT\ge3\)

\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\) ( đpcm )

Do p là nửa chu vi của tam giác

\(\Rightarrow p=\frac{a+b+c}{2}\)

Xét: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\)

\(\Leftrightarrow\frac{1}{\frac{a+b+c}{2}-a}+\frac{1}{\frac{a+b+c}{2}-b}+\frac{1}{\frac{a+b+c}{2}-c}\)

\(\Leftrightarrow\frac{1}{\frac{b+c-a}{2}}+\frac{1}{\frac{a+c-b}{2}}+\frac{1}{\frac{a+b-c}{2}}\)

\(\Leftrightarrow\frac{2}{b+c-a}+\frac{2}{a+c-b}+\frac{2}{a+b-c}\)

\(\Leftrightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\)

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với a , b > 0

\(\Rightarrow\left\{\begin{matrix}\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{2}{c}\\\frac{1}{a+c-b}+\frac{1}{a+b-c}\ge\frac{2}{a}\\\frac{1}{b+c-a}+\frac{1}{a+b-c}\ge\frac{2}{b}\end{matrix}\right.\)

Cộng theo từng vế:

\(\Rightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) ( đpcm )

Xét: \(\sqrt{\frac{a}{b+c+d}}=\frac{\sqrt{a}}{\sqrt{b+c+d}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)

\(\sqrt{\frac{b}{c+d+a}}=\frac{\sqrt{b}}{\sqrt{c+d+a}}=\frac{b}{\sqrt{b\left(c+d+a\right)}}\)

\(\sqrt{\frac{c}{d+a+b}}=\frac{\sqrt{c}}{\sqrt{d+a+b}}=\frac{c}{\sqrt{c\left(d+a+b\right)}}\)

\(\sqrt{\frac{d}{a+b+c}}=\frac{\sqrt{d}}{\sqrt{a+b+c}}=\frac{d}{\sqrt{d\left(a+b+c\right)}}\)

\(\Rightarrow VT=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\\\sqrt{b\left(c+d+a\right)}\le\frac{a+b+c+d}{2}\\\sqrt{c\left(d+a+b\right)}\le\frac{a+b+c+d}{2}\\\sqrt{d\left(a+b+c\right)}\le\frac{a+b+c+d}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a\left(b+c+d\right)}}\ge\frac{2a}{a+b+c+d}\\\frac{b}{\sqrt{b\left(c+d+a\right)}}\ge\frac{2b}{a+b+c+d}\\\frac{c}{\sqrt{c\left(d+a+b\right)}}\ge\frac{2c}{a+b+c+d}\\\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge\frac{2d}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow VT\ge\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}\)

\(\Rightarrow VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}\)

\(\Rightarrow VT\ge2\)

\(\Rightarrow\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge2\)

\(\Leftrightarrow\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\) ( đpcm )

Chiều cao tối thiểu của gương là HK

Xét tam giác OA'B'

H là trung điểm của OA'

HK // A'B'

K là trung điểm của OB'

\(\Rightarrow\) HK là đường trung bình của tam giác OA'B'

\(\Rightarrow HK=\frac{1}{2}A'B'\)

Ta có: \(AB=A'B'=1,7m\)

\(\Rightarrow HK=\frac{1}{2}.1,7m\)

\(\Rightarrow HK=0,85m\)

Vậy chiều cao tối thiểu của gương để người đó nhìn thấy ảnh mình trong gương là 0,85m

Xét: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)

\(\Leftrightarrow a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\\1+a^2\ge2\sqrt{a^2}=2a\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{ab^2}{1+b^2}\le\frac{ab^2}{2b}=\frac{ab}{2}\\\frac{bc^2}{1+c^2}\le\frac{bc^2}{2c}=\frac{bc}{2}\\\frac{ca^2}{1+a^2}\le\frac{ca^2}{2a}=\frac{ac}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}=\frac{2a-ab}{2}\\b-\frac{bc^2}{1+c^2}\ge b-\frac{bc}{2}=\frac{2b-bc}{2}\\c-\frac{ca^2}{1+a^2}\ge c-\frac{ac}{2}=\frac{2c-ac}{2}\end{matrix}\right.\)

Cộng theo từng vế:

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

\(\Rightarrow a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\ge\frac{6-\left(ab+bc+ca\right)}{2}=3-\frac{ab+bc+ca}{2}\)

Xét: \(3-\frac{ab+bc+ca}{2}\)

Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow9\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow3\ge ab+bc+ca\)

\(\Rightarrow\frac{3}{2}\ge\frac{ab+bc+ca}{2}\)

\(\Rightarrow3-\frac{3}{2}\le3-\frac{ab+bc+ca}{2}\)

\(\Rightarrow\frac{3}{2}\le3-\frac{ab+bc+ca}{2}\)

Vì \(a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\ge3-\frac{ab+bc+ca}{2}\)

\(\Rightarrow a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\) ( đpcm )

P = { 1;2;3;4;5;6 }

Q = { 3;4;5;6;7;8 }