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

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

Đặt \(\left(a;b;c\right)=\left(2\sqrt{\frac{x}{y}};2\sqrt{\frac{y}{z}};2\sqrt{\frac{z}{x}}\right)\)

\(\Rightarrow VT\ge\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}=\frac{y^2}{2xy+y^2}+\frac{z^2}{2yz+z^2}+\frac{x^2}{2zx+x^2}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Dấu "=" xảy ra khi \(a=b=c=2\)

Đặt \(\left(a;b;c\right)\rightarrow\left(\frac{2y'z'}{x'^2};\frac{2z'x'}{y'^2};\frac{2x'y'}{z'^2}\right)\) với x', y', z' > 0. Quy về chứng minh:

\(\Sigma_{cyc}\frac{x'^3}{\sqrt{x'^6+8y'^3z'^3}}\ge1\). Đặt \(\left(x'^3;y'^3;z'^3\right)=\left(x;y;z\right)\). Quy về:

\(\Sigma_{cyc}\frac{x}{\sqrt{x^2+8yz}}\ge1\). Đến đây em thấy khá quen thuộc, hình như là bài IMO nào đó, để tối lục lại.

Ok, nó đây: https://olm.vn/hoi-dap/detail/229477332481.html

Bất đắc dĩ nên em mới dùng Sigma nhiều v:

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

\(=\frac{a^2+2}{2}\).Tương tự \(\sqrt{b^3+1}\le\frac{b^2+2}{2};\sqrt{c^3+1}\le\frac{c^2+2}{2}\)

Nên \(K=\text{Σ}_{cyc}\frac{1}{\sqrt{a^3+1}}\ge\text{Σ}\frac{1}{a^2+2}=Q\)

\(Q=\frac{\text{Σ}_{cyc}[2\left(a^2+2\right)\left(b^2+2\right)]}{\text{Σ}_{cyc}\left(a^2+2\right)}\)

\(=\frac{2\left(a^2b^2+c^2b^2+a^2c^2\right)+8\left(a^2+b^2+c^2\right)+24}{2\left(a^2b^2+c^2b^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+8+a^2b^2c^2}\)

\(=\frac{2\left(\text{​​}\text{​​}\text{Σ}_{cyc}a^2b^2\right)+4\left(\text{Σ}_{cyc}a^2\right)+24+4\left(\text{Σ}_{cyc}a^2\right)}{2\left(a^2b^2+c^2b^2+a^2c^2\right)+4\left(a^2+b^2+c^2\right)+8+64}\)

\(\ge\frac{2\left(\text{Σ}a^2b^2\right)+4\left(\text{Σ}_{cyc}a^2\right)+24+12\sqrt[3]{a^2b^2c^2}}{2\left(\text{Σ}_{cyc}a^2b^2\right)+4\left(\text{Σ}_{cyc}a^2\right)+72}=1\)

Dấu "=" xảy ra khi a = b = c = 2.

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

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

Do \(abc=8\) nên tồn tại các số dương x;y;z sao cho: \(\left\{{}\begin{matrix}a=\frac{2x}{y}\\b=\frac{2y}{z}\\c=\frac{2z}{x}\end{matrix}\right.\)

\(\Rightarrow VT\ge\frac{y^2}{2x^2+y^2}+\frac{z^2}{2y^2+z^2}+\frac{x^2}{2z^2+x^2}\)

\(\Rightarrow VT\ge\frac{x^4}{x^4+2x^2z^2}+\frac{y^4}{y^4+2x^2y^2}+\frac{z^4}{z^4+2y^2z^2}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2}=1\)

Dấu "=" xảy ra khi \(a=b=c=2\)

Lời giải:
Do $abc=1$ nên đặt:

\((\sqrt{a}, \sqrt{b}, \sqrt{c})=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})\) với $x,y,z>0$

Khi đó, bài toán trở thành: Cho $x,y,z>0$. CMR:

\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}\geq 1\)

Thật vậy, áp dụng BĐT Cauchy-Schwarz:

\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}=\frac{(xz)^2}{2xyz^2+(xy)^2}+\frac{(xy)^2}{2x^2yz+(yz)^2}+\frac{(yz)^2}{2xy^2z+(xz)^2}\)

\(\geq \frac{(xz+xy+yz)^2}{2xyz^2+(xy)^2+2x^2yz+(yz)^2+2xy^2z+(xz)^2}=\frac{(xy+yz+xz)^2}{(xy+yz+xz)^2}=1\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$

thank you

3.

\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)

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

\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)

\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)

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

\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)

Đề thiếu

Dậy cho mình hỏi bài này giải sao ??

(a+!)(a^2-a+1) <= a^2+2/2 

phần còn lại tương đương <(")

(a-b)(x^2 +ab) = ( a^2 - b^2 )x 
(a-b)( x^2 +ab) = (a-b)(a+b)x
Chung a-b r đây cậu

\(a^3+b^3+1=a^3+b^3+abc\ge ab\left(a+b+c\right)\)

=>  \(\frac{\sqrt{1+a^3+b^3}}{ab}\ge\frac{\sqrt{ab\left(a+b+c\right)}}{ab}=\frac{\sqrt{a+b+c}}{\sqrt{ab}}\)

Tuong tu:  \(\frac{\sqrt{1+b^3+c^3}}{bc}\ge\frac{\sqrt{a+b+c}}{\sqrt{bc}}\)

                    \(\sqrt{1+c^3+a^3}\ge\frac{\sqrt{a+b+c}}{\sqrt{ca}}\)

suy ra:  \(\frac{\sqrt{1+a^3+b^3}}{ab}+\frac{\sqrt{1+b^3+c^3}}{bc}+\frac{\sqrt{1+c^3+a^3}}{ca}\ge\sqrt{a+b+c}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)

\(\ge\sqrt{3\sqrt[3]{abc}}.3\sqrt[3]{\frac{1}{\sqrt{ab}}.\frac{1}{\sqrt{bc}}.\frac{1}{\sqrt{ca}}}=3\sqrt{3}\)  (dpcm)