Câu 285

a) ĐKXĐ: $x\le 10.$

 \(PT\Leftrightarrow\left(\dfrac{x^3+7x^2+18x+4}{\sqrt{10-x}}-10\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\dfrac{\left(x^5+15x^4+100x^3+360x^2+740x+984\right)}{\sqrt{10-x}\left(x^3+7x^2+8x+4+10\sqrt{10-x}\right)}+1\right]=0\)

Rõ ràng biểu thức trong ngoặc vuông vô nghiệm.

Vậy $x=1$ (TMĐKXĐ)

b) Đặt $t=ab+bc+ca.$

 \(a,b,c\in\left[0,1\right]\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1.\) (1)

Từ (1) suy ra \(3abc\ge\sum c\left(a+b-1\right)=2t-\left(a+b+c\right)\ge2t-3\)

Cũng do $a,b,c\in \left[0,1\right]$ suy ra \(\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\Rightarrow abc\le\sum\left(ab-a\right)+1\)

Do đó"\(VT\le\sum\dfrac{a}{1+bc}+\sum\left(ab-a\right)+1\)

\(=\sum\left(\dfrac{a}{1+bc}-a\right)+\sum ab+1\)

\(=-abc\sum\dfrac{1}{1+bc}+ab+bc+ca+1\)

\(\le t+1-\dfrac{9abc}{t+3}\le t+1-\dfrac{3\left(2t-3\right)}{t+3}\le\dfrac{5}{2}\) 

\(\Leftrightarrow\left(2t-3\right)\left(3-t\right)\ge0\)

Do \(t\le\dfrac{\left(a+b+c\right)^2}{3}=3\) nên nếu $ab+bc+ca\ge \dfrac{3}{2}$ thì bất đẳng thức đúng.

Trong trường hợp ngược lại ta có \(VT\le t+1-\dfrac{9abc}{t+3}\le t+1\le\dfrac{3}{2}+1=\dfrac{5}{2}\) (đpcm)

Hoàn tất chứng minh.

Đẳng thức xảy ra khi (bạn đọc tự xét)

290

Ta có \(\dfrac{a^4b}{a^2+1}=a^2b-\dfrac{a^2b}{a^2+1}\ge a^2b-\dfrac{a^2b}{2a}=a^2b-\dfrac{ab}{2}\)

Chứng minh tương tự ta được:  

\(\dfrac{b^4c}{b^2+1}\ge b^2c-\dfrac{bc}{2};\dfrac{c^4a}{c^2+1}\ge c^2a-\dfrac{ca}{2}\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{ab}{2}-\dfrac{bc}{2}-\dfrac{ca}{2}\)

Áp dụng bđt Cô-si:

\(a^2b+a^2b+b^2c\ge3\sqrt[3]{a^2b\cdot a^2b\cdot b^2c}=3\sqrt[3]{a^3b^3\cdot abc}=3ab\)

Tương tự: \(b^2c+b^2c+c^2a\ge3bc;c^2a+c^2a+a^2b\ge3ca\)

\(\Rightarrow a^2b+a^2b+b^2c+b^2c+b^2c+c^2a+c^2a+c^2a+a^2b\ge3ab+3bc+3ca\Rightarrow3\left(a^2b+b^2c+c^2a\right)\ge3\left(ab+bc+ca\right)\Rightarrow a^2b+b^2c+c^2a\ge ab+bc+ca\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{1}{2}\left(ab+bc+ca\right)\ge ab+bc+ca-\dfrac{1}{2}\left(ab+bc+ca\right)=\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{3}{2}\sqrt[3]{\left(abc\right)^2}=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

C280:

Áp dụng BĐT AM-GM và BĐT BSC:

\(\dfrac{1}{\sqrt{x+3y}}+\sqrt{x+3y}\ge2\Rightarrow\dfrac{1}{\sqrt{x+3y}}\ge2-\sqrt{x+3y}\)

\(\dfrac{1}{\sqrt{y+3z}}+\sqrt{y+3z}\ge2\Rightarrow\dfrac{1}{\sqrt{y+3z}}\ge2-\sqrt{y+3z}\)

\(\dfrac{1}{\sqrt{z+3x}}+\sqrt{z+3x}\ge2\Rightarrow\dfrac{1}{\sqrt{z+3x}}\ge2-\sqrt{z+3x}\)

\(\Rightarrow P=\dfrac{1}{\sqrt{x+3y}}+\dfrac{1}{\sqrt{y+3z}}+\dfrac{1}{\sqrt{z+3x}}\)

\(\ge6-\left(\sqrt{x+3y}+\sqrt{y+3z}+\sqrt{z+3x}\right)\)

\(\ge6-\sqrt{3\left(x+3y+y+3z+z+3x\right)}\)

\(=6-\sqrt{12\left(x+y+z\right)}=3\)

\(minP=3\Leftrightarrow a=b=c=\dfrac{1}{4}\)

Bài 7) 

\(bđt\Leftrightarrow4\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow a^3+b^3+c^3+4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)+6abc\)\(\Leftrightarrow ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\ge6abc\)

\(\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

(Đúng theo Cô Si)

"=" khi a=b=c=1

Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)

Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)

DM CHƯA HỌC ĐẾN

Bài nào đó k ghi số nên không bt gọi ntn:

Chuẩn hóa x + y + z = 3. Ta cần cm \(x^2y+y^2z+z^2x+xyz\le4\).

Giả sử \(z=mid\left\{x,y,z\right\}\Rightarrow\left(x-z\right)\left(y-z\right)\le0\)

\(\Leftrightarrow xy+z^2\le xz+yz\)

\(\Leftrightarrow x^2y+xz^2\le x^2z+xyz\).

Từ đó \(x^2y+y^2z+z^2x+xyz\le x^2z+xyz+y^2z+xyz=z\left(x+y\right)^2\le\dfrac{\dfrac{\left(2z+x+y+x+y\right)^3}{27}}{2}=4\).

Câu cuối:

Áp dụng BĐT BSC:

\(\dfrac{a}{\sqrt{a^2+b+c}}=\sqrt{\dfrac{a^2}{a^2+b+c}}=\sqrt{\dfrac{a^2\left(1+b+c\right)}{\left(a^2+b+c\right)\left(1+b+c\right)}}\le\sqrt{\dfrac{a^2\left(1+b+c\right)}{\left(a+b+c\right)^2}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)

Tương tự \(\dfrac{b}{\sqrt{b^2+c+a}}=\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\); \(\dfrac{c}{\sqrt{c^2+a+b}}=\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)

Khi đó \(VT\le\Sigma\left(\dfrac{a}{a+b+c}.\sqrt{1+b+c}\right)\)

Giả sử \(a\ge b\ge c\)

Áp dụng BĐT Chebyshev với bộ \(\dfrac{a}{a+b+c};\dfrac{b}{a+b+c};\dfrac{c}{a+b+c}\) và \(\sqrt{1+b+c};\sqrt{1+c+a};\sqrt{1+a+b}\):

\(VT\le\dfrac{1}{3}\Sigma\dfrac{a}{a+b+c}.\Sigma\sqrt{1+a+b}=\dfrac{\Sigma\sqrt{1+a+b}}{3}\)

\(\le\dfrac{\sqrt{3\left(3+2a+2b+2c\right)}}{3}\)

\(\le\dfrac{\sqrt{9+6\sqrt{3\left(a^2+b^2+c^2\right)}}}{3}=\sqrt{3}\)

Đẳng thức xảy ra khi \(a=b=c=1\)

axb=2xawfd458uf

Câu 5 em thấy thầy làm từ chiều, em nghĩ anh nên đổi câu khác:

Cho \(x,y,z\ge0\).Tìm giá trị lớn nhất :\(P=\dfrac{x}{x^2 y^2 2} \dfrac{y}{y^2 z^2 2} \dfrac{z}{z^2 x^2 2}\) - Hoc24

Câu 266 là >= chứ nhỉ?

2.

\(\left(a+b\right)^2\ge4ab\ge16\Rightarrow a+b\ge4\)

\(\dfrac{a^2+b^2}{a+b}\ge\dfrac{\left(a+b\right)^2}{2\left(a+b\right)}=\dfrac{a+b}{2}\)

Nên ta chỉ cần chứng minh: \(\dfrac{a+b}{2}\ge\dfrac{6}{a+b-1}\)

\(\Leftrightarrow\left(a+b\right)\left(a+b-1\right)-12\ge0\)

\(\Leftrightarrow\left(a+b-4\right)\left(a+b+3\right)\ge0\) (luôn đúng với mọi \(a+b\ge4\))

Dấu "=" xảy ra khi \(a=b=2\)

Câu cuối:

Ta chứng minh BĐT phụ sau: với mọi x;y;z dương, ta luôn có: \(\dfrac{x^3+y^3}{x^2+y^2}\ge\dfrac{x+y}{2}\)

Thật vậy, bất đẳng thức tương đương:

\(2\left(x^3+y^3\right)\ge\left(x+y\right)\left(x^2+y^2\right)\)

\(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (đúng)

Áp dụng:

\(P\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\ge6\)

\(P_{min}=6\) khi \(a=b=c=2\)

\(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì a>b>0 =>a²+ab+3b²>0 nên từ (1) ta có a=2b

Vậy biểu thức \(A=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)