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Khác số chút thoyy.

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 \(a=b=c=1\)

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

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

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

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

\(=\left(a+b+c-3\right)^2+2002\)

đúng rồi đấy

a 2 + b 2 + c 2 = a + b + c 2 − 2 ab + bc + ca = a + b + c 2 − 6. P = a + b + c 2 − 6 − 6 a + b + c + 2017 = a + b + c 2 − 6 a + b + c + 9 + 2002 = a + b + c − 3 2 + 2002 Mà a + b + c − 3 2 ≥ 0nên GTNN của P bằng 2002

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 \(a=b=c=1\)

Hi vọng là tìm GTLN:

Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).

Áp dụng bất đẳng thức AM - GM ta có: 

\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)

\(\Rightarrow a+b+c\le3\).

Áp dụng bất đẳng thức Schwarz ta có:

\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).

Đẳng thức xảy ra khi a = b = c = 1.

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+5=11\)

Vậy: P(min)=11  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)