\(M=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

\(M\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{ab+bc+ca}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow M\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}=9+\dfrac{7.3}{\left(a+b+c\right)^2}=9+21=30\)

\(Min_M=30\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo

\(m\text{≥}\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)

\(=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

≥ \(\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)\(+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

CM BĐT: \(a^2+b^2+c^2\text{≥}ab+bc+ca\)

⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\text{≥}0\) (luôn đúng)

⇒ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ca\right)\)

⇒ \(\dfrac{\left(a+b+c\right)^2}{3}\text{≥}ab+bc+ca\)

⇒ \(m\text{≥}\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=9+21=30\) 

(vì a+b+c=1)

Vậy...

Lời giải:

\(Q=\frac{ab}{c+ab}+\frac{ac}{b+ac}+\frac{bc}{a+bc}-\frac{1}{4abc}=\frac{ab}{c(a+b+c)+ab}+\frac{ac}{b(a+b+c)+ac}+\frac{bc}{a(a+b+c)+bc}-\frac{1}{4abc}\)

\(=\frac{ab}{(c+a)(c+b)}+\frac{ac}{(b+a)(b+c)}+\frac{bc}{(a+b)(a+c)}-\frac{1}{4abc}\)

\(=\frac{ab(a+b)+ac(a+c)+bc(b+c)}{(a+b)(b+c)(c+a)}-\frac{1}{4abc}\)

\(=\frac{(a+b)(b+c)(c+a)-2abc}{(a+b)(b+c)(c+a)}-\frac{1}{4abc}\) (đẳng thức quen thuộc \((a+b)(b+c)(c+a)=ab(a+b)+bc(b+c)+ca(c+a)+2abc\) )

\(=1-\left(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{4abc}\right)\)

Áp dụng BĐT AM-GM:

\(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{108abc}\geq 2\sqrt{\frac{1}{54(a+b)(b+c)(c+a)}}\).

Mà \(2=(a+b)+(b+c)+(c+a)\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}\Rightarrow (a+b)(b+c)(c+a)\leq \frac{8}{27}\)

\(\Rightarrow \frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{108abc}\geq \frac{1}{2}\)

\(1=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{27}\)

\(\Rightarrow \frac{13}{54abc}\geq \frac{13}{2}\)

Do đó: \(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{4abc}\geq 7\)

\(\Rightarrow Q\leq 1-7=-6=Q_{\max}\)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

nhân cả vế với abc ta có điều cần chứng minh

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

VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)

=>(đpcm)

mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\) 

chỉ cần chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé

nhân cả vế với abc ta có điều cần chứng minh

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

VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)

=>(đpcm)

mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)

chỉ cần chứng minh được\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+x}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé.

o boai

Áp dụng BĐT Cô-si:

\(A\le\dfrac{a+b}{2\sqrt{c+ab}}+\dfrac{b+c}{2\sqrt{a+bc}}+\dfrac{c+a}{2\sqrt{b+ac}}\)\(\le\dfrac{a+b}{2\sqrt{2\sqrt{abc}}}+\dfrac{b+c}{2\sqrt{2\sqrt{abc}}}+\dfrac{c+a}{2\sqrt{2\sqrt{abc}}}\)\(=\dfrac{a+b+c}{\sqrt[4]{4abc}}=\dfrac{1}{\sqrt[4]{4abc}}\ge\dfrac{1}{\sqrt{\left(a+b+c\right).\dfrac{2}{3}}}\)(BĐT Cô-si)\(=\dfrac{1}{\sqrt{\dfrac{2}{3}}}=\dfrac{\sqrt{6}}{2}\)

Vậy A_min=\(\dfrac{\sqrt{6}}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)

\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)

\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)

Áp dụng bđt Cauchy ta có :

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)

Thiết lập tương tự và thu lại ta có :

\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)

Dấu '' = '' xảy ra khi \(a=b=c=3\)

b)Ta có: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c\left(1\right)\)

\(\Leftrightarrow\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4}{abc}\ge a+b+c\)

\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta xét BĐT phụ: \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

Cộng các BĐT phụ vừa chứng minh:

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)

Áp dụng vào bài, ta có:

\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)

Áp dụng lần nữa:

\(a^2b^2+b^2c^2+c^2a^2\ge ab^2c+bc^2a+a^2bc=abc\left(a+b+c\right)\)

Vậy ta suy ra được điều phải chứng minh

a) Đặt vế trái BĐT là P

\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)8.8}}=\dfrac{3a}{4}\)

Tương tự: \(\dfrac{b^3}{\left(1+a\right)\left(1+c\right)}+\dfrac{1+a}{8}+\dfrac{1+c}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế theo vế các BĐT vừa chứng minh

\(P+\dfrac{6+2a+2b+2c}{8}\ge\dfrac{3a+3b+3c}{4}\)

\(P\ge\dfrac{3a+3b+3c}{4}-\dfrac{2\left(3+a+b+c\right)}{8}=\dfrac{3a+3b+3c-a-b-c-3}{4}=\dfrac{2\left(a+b+c\right)-3}{4}\)

\(a+b+c\ge3\sqrt[3]{abc}=3\)

\(\Rightarrow P\ge\dfrac{2.3-3}{4}=\dfrac{3}{4}\)

Câu c) là Cm \(\le\)1 ạ, Mìk thiếu đề