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

Thiết lập tương tự: \(\frac{1}{\sqrt{b^3+1}}\ge\frac{2}{b^2+2}\) ; \(\frac{1}{\sqrt{c^3+1}}\ge\frac{2}{c^2+2}\)

\(\Rightarrow VT\ge\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}=\frac{1}{\frac{a^2}{2}+1}+\frac{1}{\frac{b^2}{2}+1}+\frac{1}{\frac{c^2}{2}+1}\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=\frac{1}{8}\)

\(\Rightarrow VT\ge\frac{x^2}{x^2+\frac{1}{2}}+\frac{y^2}{y^2+\frac{1}{2}}+\frac{z^2}{z^2+\frac{1}{2}}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+\frac{3}{2}}\)

\(\Rightarrow VT\ge\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x^2+y^2+z^2+\frac{3}{2}}\ge\frac{x^2+y^2+z^2+6.\sqrt[3]{\left(xyz\right)^2}}{x^2+y^2+z^2+\frac{3}{2}}=\frac{x^2+y^2+z^2+\frac{3}{2}}{x^2+y^2+z^2+\frac{3}{2}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) hay \(a=b=c=2\)

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

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

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.

Lời giải:
Áp dụng BĐT AM-GM:

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

\(b^3+1\leq \left(\frac{b^2+2}{2}\right)^2\)

\(\Rightarrow \sqrt{(a^3+1)(b^3+1)}\leq \frac{(a^2+2)(b^2+2)}{4}\)

\(\Rightarrow \frac{a^2}{\sqrt{(a^3+1)(b^3+1)}}\geq \frac{4a^2}{(a^2+2)(b^2+2)}\)

Hoàn toàn tương tự với các phân thức còn lại:

\(\Rightarrow \text{VT}\geq \underbrace{\frac{4a^2}{(a^2+2)(b^2+2)}+\frac{4b^2}{(b^2+2)(c^2+2)}+\frac{4c^2}{(c^2+2)(a^2+2)}}_{M}\)

Ta cần CM \(M\geq \frac{4}{3}\)

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

\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (a^2+2)(b^2+2)(c^2+2)\)

\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (abc)^2+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2+b^2+c^2)+8\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2(a^2+b^2+c^2)\geq 72\)

Điều này luôn đúng do theo BĐT AM-GM thì: \(\left\{\begin{matrix} a^2b^2+b^2c^2+c^2a^2\geq 3\sqrt[3]{(abc)^4}=3\sqrt[3]{8^4}=48\\ 2(a^2+b^2+c^2)\geq 6\sqrt[3]{(abc)^2}=6\sqrt[3]{8^2}=24\end{matrix}\right.\)

Do đó ta có đpcm

Dấu "=" xảy ra khi $a=b=c=2$

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)

1.

\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)

\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)

Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá

2.

\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

Đặt \(x+y+z=t\Rightarrow0< t\le1\)

\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

3.

\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)

Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)

Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)

4.

ĐKXĐ: \(-2\le x\le2\)

\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)

\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)

Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)

\(y_{min}=-2\) khi \(x=-2\)

5.

\(\frac{a\sqrt{b-1}+b\sqrt{a-1}}{ab}=\frac{1.\sqrt{b-1}}{b}+\frac{1.\sqrt{a-1}}{a}\le\frac{1+b-1}{2b}+\frac{1+a-1}{2a}=1\)

\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

Dấu "=" xảy ra khi \(a=b=2\)

6. Áp dụng BĐT cơ bản:

\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow\left(ab+bc+ca\right)^2\ge3\left(ab.bc+bc.ca+ab+ca\right)\)

\(\Rightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

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

\(P\ge\frac{2a}{b^2+4}+\frac{2b}{c^2+4}+\frac{2c}{a^2+4}\)

\(2P\ge\frac{4a}{b^2+4}+\frac{4b}{c^2+4}+\frac{4c}{a^2+4}=a-\frac{ab^2}{b^2+4}+b-\frac{bc^2}{c^2+4}+a-\frac{ca^2}{a^2+4}\)

\(2p\ge a+b+c-\left(\frac{ab^2}{4b}+\frac{bc^2}{4c}+\frac{ca^2}{4a}\right)\)

\(2P\ge6-\frac{1}{4}\left(ab+bc+ca\right)\ge6-\frac{1}{12}\left(a+b+c\right)^2=3\)

\(\Rightarrow P\ge\frac{3}{2}\)

Dấu " = " xảy ra khi \(a=b=c=2\)