a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)

b) Ta có   \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)

\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)

Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

Lời giải:

Ta thấy:

\(\text{VT}=\frac{c^2}{2ab^2c^2+c^2}+\frac{a^2}{2bc^2a^2+a^2}+\frac{b^2}{2ca^2b^2+b^2}\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}(2ab^2c^2+c^2+2bc^2a^2+a^2+2ca^2b^2+b^2)\geq (c+a+b)^2\)

\(\Leftrightarrow \text{VT}\geq \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}(*)\)

Áp dụng BĐT Am-GM:

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

\(\Rightarrow 2abc(ab+bc+ac)\leq 2(ab+bc+ac)\)

\(\Rightarrow \frac{(a+b+c)^2}{2abc(ab+bc+ac)+a^2+b^2+c^2}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)+a^2+b^2+c^2}=1(**)\)

Từ \((*); (**)\Rightarrow \text{VT}\geq 1\)

Ta có đpcm. Dấu "=" xảy ra khi $a=b=c=1$

Cách khác bằng AM-GM:

\(\text{VT}=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)(1)\)

Áp dụng BĐT AM-GM:

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\leq \frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^2a^4}}=\frac{2}{3}(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2})\)

\(\leq \frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)=\frac{2}{3}(a+b+c)=2(2)\)

Từ \((1);(2)\Rightarrow \text{VT}\geq 3-2=1\) (đpcm)

dùng bđt bunhia nha

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)

\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)

\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)

     Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)

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

\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)

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

Sau đó bạn thực hiện tiếp nhé.

Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)

Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)

Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)

Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)

Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)

Áp dụng bđt svac-xơ có:

\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\)

<=> \(A\ge\frac{9}{\left(a+b+c\right)^2}\)

Với a,b,c>0 và a+b+c \(\le1\) => 0<(a+b+c)²\(\le1\)=> \(\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)

=>A\(\ge9\)

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

ta có A\(\ge\frac{9}{\left(a+b+c\right)^2}=9\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Cách : AM - GM :

\(VT=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)\left(1\right)\)

Áp dụng BĐT AM - GM :

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\le\frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^aa^4}}=\frac{2}{3}\left(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2}\right)\)

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

Từ (1) và (2) \(\Rightarrow VT\ge3-2=1\left(đpcm\right)\)

          \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\)\(\frac{ab+bc+ca}{abc}=0\)

\(\Rightarrow\)\(ab+bc+ca=0\)

\(\Rightarrow\)\(\hept{\begin{cases}ab=-\left(bc+ca\right)\\bc=-\left(ab+ca\right)\\ca=-\left(ab+bc\right)\end{cases}}\)

\(\Rightarrow\)\(\hept{\begin{cases}a^2+2bc=a^2+bc-ab-ca=\left(a-b\right)\left(a-c\right)\\b^2+2ac=b^2+ac-ab-bc=\left(b-c\right)\left(b-a\right)\\c^2+2ab=c^2+ab-bc-ca=\left(c-a\right)\left(c-b\right)\end{cases}}\)

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

P/S: đến đây tự lm nhé