Áp dụng BĐT Bunhiacopxki:

\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}\le\sqrt{2\left(\frac{2}{a}+\frac{2}{b}\right)}=2\sqrt{\frac{a+b}{ab}}\)

Tương tự: \(\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le2\sqrt{\frac{b+c}{bc}}\) ; \(\sqrt{\frac{2}{c}}+\sqrt{\frac{2}{a}}\le2\sqrt{\frac{c+a}{ca}}\)

Cộng vế với vế ta sẽ có điều phải chứng minh

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)

Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)

Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)

=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)

=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)

Ta có đpcm

https://olm.vn/thanhvien/chibiverycute là con chó

\(L.H.S=\Sigma_{cyc}\frac{a^2}{b}=\Sigma_{cyc}\left(\frac{a^2}{b}-a+b\right)=\Sigma_{cyc}\frac{a^2-ab+b^2}{b}\)

\(=\Sigma_{cyc}\left(\frac{a^2-ab+b^2}{b}+b\right)-\left(a+b+c\right)\)

\(\ge2\Sigma_{cyc}\sqrt{a^2-ab+b^2}-\left(a+b+c\right)\)

\(=\Sigma_{cyc}\sqrt{a^2-ab+b^2}+\Sigma_{cyc}\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}-\left(a+b+c\right)\)

\(\ge\Sigma_{cyc}\sqrt{a^2-ab+b^2}+\Sigma_{cyc}\sqrt{\frac{1}{4}\left(a+b\right)^2}-\left(a+b+c\right)=\Sigma_{cyc}\sqrt{a^2-ab+b^2}=R.H.S\)

Đẳng thức xảy ra khi a = b = c

SOS ảo diệu hơn!

\(VT-VP=\Sigma_{cyc}\left[\frac{a^2}{b^2\left(\sqrt{\frac{a^2-ab+b^2}{b}}+\sqrt{b}\right)^2}^2+\Sigma_{cyc}\frac{3}{4\sqrt{a^2-ab+b^2}+2\left(a+b\right)}\right]\left(a-b\right)^2\ge0\) (đúng)

P/s: Nếu rảnh thì check hộ em:P

https://goo.gl/BjYiDy

Với \(a,b,c\ge0\). Khi đó ta có

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

Chứng minh: \(\left(ab+bc+ca\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a^2+b^2+c^2+abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge a^2+b^2+c^2\)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{ab+bc+ac}\)

Với \(a,b,c\ge0\) ta có

\(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(b+a\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(c+a\right)}}\ge1\)

Áp dụng bất đẳng thức AM-GM ta có:

\(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}=\Sigma\sqrt{\frac{ab\left(2ab+2bc+2ac\right)^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}\)

\(\ge\Sigma\sqrt{\frac{ab\left[a\left(b+c\right)+b\left(a+c\right)\right]^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ac\right)^2}}\)

\(\ge\Sigma\sqrt{\frac{ab.4a\left(b+c\right)b\left(a+c\right)}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}=\Sigma\frac{ab}{ab+bc+ca}\)

Từ đó ta có \(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\frac{ab+bc+ca}{ab+bc+ca}=1\)

chứng minh bài toán:

Đặt \(\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ac}}=t\ge1\)

Ta có: \(\left(\Sigma\sqrt{\frac{a}{b+c}}\right)^2=\Sigma\frac{a}{b+c}+2\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\frac{a^2+b^2+c^2}{ab+bc+ac}+2=t^2+2\)

Từ đây ta chứng minh \(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}\ge\frac{7\sqrt{2}}{2}\)

Áp dụng bất đẳng thức bunhiacopxki ta có:

\(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}=\frac{\sqrt{\left(t^2+2\right)\left(6+2\right)}}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}\ge\frac{t\sqrt{6}+2}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}=\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\)

Áp dụng bất đẳng thức Cauchy ta đc:

\(\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\ge3\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}\)

Vậy ta có đpcm

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

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

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

Tham khảo

Câu hỏi của Châu Trần - Toán lớp 9 - Học toán với OnlineMath

à xl gửi lộn

Oh yeah mik lm đc r.

\(\frac{1}{\sqrt{ab+a+2}}< =\frac{1}{ab+a+2}+\frac{1}{4}\\ \)

\(=>VT< =sigma\frac{1}{ab+a+2}+\frac{3}{4}\)

\(Có\frac{1}{ab+a+2}< =\frac{1}{4}\left(\frac{1}{ab+1}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

\(CMTT\frac{1}{bc+c+2}< =\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{c+1}\right)\)

\(\frac{1}{ca+c+2}< =\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\)

Cộng lại => Vế trái <= 1/4.3/4+3/4=3/2

=> đpcm.