\(\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\ge16\)

Ap dung bdt amgm va bdt bunhiacpoxki taok:

\(VT=\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\)

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

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

\(=\left(a+b+c+3\right)\left(\frac{1}{a+b+c}+3\right)\)

\(\ge\left(1+3\right)^2=16=VP\)

dùng bất đẳng thức cô-si và bất đẳng thức bu-nhi-a-cop-xki

VT=\(\frac{2}{2a+b+2\sqrt{2bc}}+3\)\(\ge\frac{2}{2a+2b+2c}+3\)=\(\frac{1}{a+b+c}+\frac{9}{3}\)

Áp dụng bđt cauchy-schwarzta được:

VT\(\ge\frac{\left(1+3\right)^2}{a+b+c+3}=\frac{6}{a+b+c+3}\)

ta cần chứng minh: a+b+c\(\le\sqrt{2b^2+2\left(a+c\right)^2}\)

Thật vậy ta có

\(2b^2+2\left(a+c\right)^2-\left(a+b+c\right)^2=b^2+a^2+c^2+2ac-2bc-2ab=\left(b-a-c\right)^2\ge0\)Suy ra a+b+c\(\le\sqrt{2b^2+2\left(a+c\right)^2}\)

vậy VT\(\ge VP\)

Sai đề ở vế phải. Cái này tôi làm rồi nên biết:  819598 (học 24)

BDT cần cm tương đương

\(\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\ge16\)

Áp dụng bdt C-S và AM-GM:

\(VT=\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\)

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

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

\(=\left(a+b+c+3\right)\left(\frac{1}{a+b+c}+3\right)\)

\(\ge\left(3+1\right)^2=16=VP\)

dau '=' khi a+b+c=1, b=a+c, 2c=b bn tự giải not

Chuyên toán Vĩnh Phúc đây mà :) Em chụp lại nha,chớ e mà viết ra nhiều người nhảy vào cà khịa ghê lắm:(

Viết BĐT về dạng \(\frac{2}{2a+b+2\sqrt{2bc}}-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\ge0\)

Ta có: \(\frac{2}{2a+b+2\sqrt{2bc}}\ge\frac{2}{2a+b+b+2c}=\frac{1}{a+b+c}\)

Đẳng thức xảy ra <=> b=2c

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a+b+c\right)^2\le\left(1+1\right)\left[\left(a+c\right)^2+b^2\right]\)

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

\(\Rightarrow-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\ge-\frac{16}{a+b+c+3}\)

Đẳng thức xảy ra <=> a+c=b

\(\Rightarrow\frac{2}{2a+b+2\sqrt{bc}}-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}+3\ge\frac{1}{a+b+c}-\frac{16}{a+b+c+3}+3\)

\(=\frac{3\left(a+b+c-1\right)^2}{\left(a+b+c\right)\left(a+b+c+3\right)}\ge0\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b+c-1=0\\b=2c\\a+c=b\end{cases}\Leftrightarrow\hept{\begin{cases}a=c=\frac{1}{4}\\b=\frac{1}{2}\end{cases}}}\)

\(4\left(a+b+c\right)=a^2+\left(b+c\right)^2\ge\dfrac{1}{2}\left(a+b+c\right)^2\)

\(\Rightarrow a+b+c\le8\)

\(a^2+16-16\ge8a-16\)

\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{8100}{\sqrt{2a+2b+1}+\sqrt{2c+1}}\)

\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{48600}{6\sqrt{2a+2b+1}+6\sqrt{2c+1}}\)

\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{24300}{a+b+c+10}\)

\(\Rightarrow P\ge8\left(a+b+c+10+\dfrac{324}{a+b+c+10}\right)+\dfrac{21708}{a+b+c+10}-96\)

\(\Rightarrow P\ge16.\sqrt{324}+\dfrac{21708}{18}-96=1398\)

Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(4;0;4\right)\)

\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)

Ta có:

\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)

\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng BĐT Cauchy-Schwarz:

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

TK: Cho các số thực dương a, b, c thỏa mãn a + b+ c = 3. Chứng minh rằng: \(\sqrt{2a^2+\frac{7}{b^2}}+\sqrt{2b^2+\frac{7}{... - Hoc24