- Theo BĐT Cauchy ta có:

\(\sqrt{a.1}\le\dfrac{a+1}{2}\)

\(\sqrt{b.1}\le\dfrac{b+1}{2}\)

\(\sqrt{c.1}\le\dfrac{c+1}{2}\)

\(\sqrt{ab}\le\dfrac{a+b}{2}\)

\(\sqrt{bc}\le\dfrac{b+c}{2}\)

\(\sqrt{ca}\le\dfrac{c+a}{2}\)

\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)

\(\Rightarrow a=b=c=1\)

\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)

Ta có:

\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)\ge0\)

\(\Rightarrow ab+ac+bc\le\dfrac{2.3}{2}=3\) (1)

Lại có: \(a^2+1+b^2+1+c^2+1\ge2a+2b+2c\)

\(\Rightarrow a+b+c\le\dfrac{a^2+b^2+c^2+3}{2}=3\) (2)

Cộng vế với vế của (1) và (2) ta được:

\(a+b+c+ab+ac+bc\le6\)

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

\(\Rightarrow A=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2017}}=\dfrac{3}{3}=1\)

Ta có:

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

Thiết lập 2 BĐT tương tự ta có:

\(b^2+c^2\ge2bc;c^2+a^2\ge2ca\)

Và \(\left(a-1\right)^2\ge0\Leftrightarrow a^2-2a+1\Leftrightarrow a^2+1\ge2a\)

Và tương tự \(b^2+1\ge2b;c^2+1\ge2c\)

Cộng theo vế các BĐT trên ta có:

\(2ab+2bc+2ca+2a+2b+2c\le3a^2+3b^2+3c^2+3\)

\(\Leftrightarrow2\left(ab+bc+ca+a+b+c\right)\le3\left(a^2+b^2+c^2+1\right)\)

\(\Leftrightarrow2\left(ab+bc+ca+a+b+c\right)\le12\)

\(\Leftrightarrow ab+bc+ca+a+b+c\le6\)

Đẳng thức xảy ra khi \(a=b=c=1\)

Khi đó \(A=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2014}}=\dfrac{1+1+1}{1+1+1}=1\)

Lời giải:

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow \frac{abc}{c(a+b)}=\frac{abc}{a(b+c)}=\frac{bca}{b(c+a)}\)

\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)

\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)

\(\Rightarrow a=b=c\) (do $a,b,c>0$)

$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(P=\dfrac{9}{ab+bc+ca}+\dfrac{2}{a^2+b^2+c^2}\)

\(=2\left[\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\right]+\dfrac{5}{ab+bc+ca}\)

\(\ge2.\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\dfrac{5}{ab+bc+ca}\)

\(=\dfrac{18}{1}+\dfrac{5}{ab+bc+ca}\ge18+5.\dfrac{3}{\left(a+b+c\right)^2}=18+15=33\)

Đẳng thức xảy ra khi a=b=c=1/3.

Vậy GTNN của P là 33.

áp dụng bất đẳng thức phụ \(\dfrac{1}{a}+\dfrac{1}{b}\)≥\(\dfrac{4}{a+b}\)<=>(a-b)²≥0 (luôn đúng)
Ta có P≥\(\dfrac{\left(3+\sqrt{2}\right)^2}{\left(a+b+c\right)^2}\)=(3+\(\sqrt{2}\))²
Dấu = xảy ra <=> a=b=c=1/3