Ta có: \(\frac{a}{1+4b^2}=\frac{a\left(1+4b^2\right)-4ab^2}{1+4b^2}=a-\frac{4ab^2}{1+4b^2}\ge a-\frac{4ab^2}{2\sqrt{4b^2.1}}=a-\frac{2ab^2}{2b}=a-ab\)(bđt cosi)

CMTT: \(\frac{b}{1+4a^2}\ge b-ab\)

=> P \(\ge a+b-2ab=4ab-2ab=2ab\)

Mặt khác ta có: \(a+b\ge2\sqrt{ab}\)(cosi)

=> \(4ab\ge2\sqrt{ab}\) <=> \(2ab\ge\sqrt{ab}\)<=> \(4a^2b^2-ab\ge0\) <=> \(ab\left(4ab-1\right)\ge0\)

<=> \(\orbr{\begin{cases}ab\le0\left(loại\right)\\ab\ge\frac{1}{4}\end{cases}}\)(vì a,b là số thực dương)

=> P \(\ge2\cdot\frac{1}{4}=\frac{1}{2}\)

Dấu "=" xảy ra <=> a = b = 1/2

Vậy MinP = 1/2 <=> a = b= 1/2

Ta có: \(a+b=4ab\le\left(a+b\right)^2\Leftrightarrow\left(a+b\right)\left[\left(a+b\right)-1\right]\ge0\)

Mà \(a+b>0\Rightarrow a+b\ge1\)

Áp dụng BĐT Cô-si, ta có: \(P=\frac{a}{1+4b^2}+\frac{b}{1+4a^2}=\left(a-\frac{4ab^2}{1+4b^2}\right)+\left(b-\frac{4a^2b}{1+4a^2}\right)\)\(\ge\left(a-\frac{4ab^2}{4b}\right)+\left(b-\frac{4a^2b}{4a}\right)=\left(a+b\right)-2ab=\left(a+b\right)-\frac{a+b}{2}=\frac{a+b}{2}\ge\frac{1}{2}\)

Đẳng thức xảy ra khi a = b = ¹/₂

\(\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{9^2}{9+3}=\dfrac{27}{4}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Chứng minh BĐT \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\) với \(\left(a,b,c>0\right)\)

Trước hết ta cm \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)\(\Leftrightarrow\frac{x^2b+y^2a}{ab}\ge\frac{x^2+y^2+2xy}{a+b}\)\(\Leftrightarrow\left(x^2b+y^2a\right)\left(a+b\right)\ge ab\left(x^2+y^2+2xy\right)\)(vì tất cả các tử số và mẫu số đều dương)

\(\Leftrightarrow x^2ab+y^2ab+x^2b^2+y^2a^2\ge abx^2+aby^2+2abxy\)\(\Leftrightarrow x^2b^2-2abxy+y^2a^2\ge0\)\(\Leftrightarrow\left(xb-ya\right)^2\ge0\)(luôn đúng)

Vậy BĐT được cm 

Để có đpcm thì ta chỉ cần áp dụng 2 lần BĐT ta vừa chứng minh xong:

\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y\right)^2}{a+b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Nma mik lớp 4

Áp dụng BĐT Svácxơ, ta có:

\(\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{81}{12}=\dfrac{27}{4}\)

Dấu "=" ⇔ a=b=c=3

Áp dụng BĐT Cô-si:

\(\dfrac{a^2}{b+1}+\dfrac{9}{16}\left(b+1\right)\ge2\sqrt{\dfrac{9a^2\left(b+1\right)}{16\left(b+1\right)}}=\dfrac{3a}{2}\) 

Tương tự: \(\dfrac{b^2}{c+1}+\dfrac{9}{16}\left(c+1\right)\ge\dfrac{3b}{2}\) ; \(\dfrac{c^2}{a+1}+\dfrac{9}{16}\left(a+1\right)\ge\dfrac{3c}{2}\)

Cộng vế:

\(VT+\dfrac{9}{16}\left(a+b+c+3\right)\ge\dfrac{3}{2}\left(a+b+c\right)\)

\(\Leftrightarrow VT+\dfrac{27}{4}\ge\dfrac{27}{2}\Rightarrow VT\ge\dfrac{27}{4}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Đặt: \(A=\sqrt{a^2+\frac{1}{a^2}}+\sqrt{b^2+\frac{1}{b^2}}+\sqrt{c^2+\frac{1}{c^2}}\), khi đó ta được:

\(A^2=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)

\(+2\cdot\sqrt{\left(a^2+\frac{1}{a^2}\right)\left(b^2+\frac{1}{b^2}\right)}+2\cdot\sqrt{\left(b^2+\frac{1}{b^2}\right)\left(c^2+\frac{1}{c^2}\right)}+2\cdot\sqrt{\left(c^2+\frac{1}{c^2}\right)\left(a^2+\frac{1}{a^2}\right)}\)

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(\sqrt{\left(a^2+\frac{1}{a^2}\right)\left(b^2+\frac{1}{b^2}\right)}\ge\sqrt{\left(ab+\frac{1}{ab}\right)^2}=ab+\frac{1}{ab}\)

\(\sqrt{\left(b^2+\frac{1}{b^2}\right)\left(c^2+\frac{1}{c^2}\right)}\ge\sqrt{\left(bc-\frac{1}{bc}\right)^2}=bc+\frac{1}{bc}\)

\(\sqrt{\left(c^2+\frac{1}{c^2}\right)\left(a^2+\frac{1}{a^2}\right)}\ge\sqrt{\left(ca+\frac{1}{ca}\right)^2}=ca+\frac{1}{ca}\)

Do đó ta có:

\(A^2\ge a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(=\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\ge\left(a+b+c\right)^2+\left(\frac{9}{a+b+c}\right)^2=82\)

Hay \(A\ge\sqrt{82}\), vậy bất đẳng thức được chứng minh.

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)