Giải casio được không?/

Với \(x\in\left(0;1\right)\) thì luôn có \(x^{\frac{1}{2}}< x^{\frac{1}{3}}\Leftrightarrow\sqrt{x}< \sqrt[3]{x}\)

Hay \(\sqrt{abc}< \sqrt[3]{abc}\). Áp dụng BĐT AM-GM ta có:

\(\sqrt{abc}< \sqrt[3]{abc}\le\frac{a+b+c}{3}\)

\(\sqrt{\left(1-a\right)\left(1-b\right)\left(1-c\right)}< \sqrt[3]{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)

\(\le\frac{\left(1-a\right)+\left(1-b\right)+\left(1-c\right)}{3}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT< \frac{a+b+c+1-a+1-b+1-c}{3}=1\)

Ta có tính chất: \(\sqrt{x+y}< \sqrt{x}+\sqrt{y}\left(x,y>0\right)\) 

Thật vậy, với x, y > 0, ta có: \(\left(\sqrt{x}+\sqrt{y}\right)^2=x+y+2\sqrt{xy}>x+y\)

\(\Rightarrow\sqrt{x}+\sqrt{y}>\sqrt{x+y}\)

Áp dụng BĐT Cauchy-Schwarz và sử dụng tính chất trên, ta được: \(\left(\sqrt{abc}+\sqrt{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\right)^2\)\(=\left(\sqrt{a}.\sqrt{bc}+\sqrt{1-a}.\sqrt{\left(1-b\right)\left(1-c\right)}\right)^2\)\(\le\left[a+\left(1-a\right)\right]\left[bc+\left(1-b\right)\left(1-c\right)\right]=bc+\left(1-b\right)\left(1-c\right)\)

\(\Rightarrow\sqrt{abc}+\sqrt{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\le\sqrt{bc+\left(1-b\right)\left(1-c\right)}\)\(< \sqrt{bc}+\sqrt{\left(1-b\right)\left(1-c\right)}\)(1)

Mặt khác: \(\left(\sqrt{bc}+\sqrt{\left(1-b\right)\left(1-c\right)}\right)^2\le\left[b+\left(1-b\right)\right]\left[c+\left(1-c\right)\right]\)\(=1\)

\(\Rightarrow\sqrt{bc}+\sqrt{\left(1-b\right)\left(1-c\right)}\le1\)(2)

Từ (1) và (2) suy ra\(\sqrt{abc}+\sqrt{\left(1-a\right)\left(1-b\right)\left(1-c\right)}< 1\left(q.e.d\right)\)

ko biết

không bt làm thì đừng có mak trả lời

Ta có:

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

\(\le\frac{1}{4\left(a+b\right)\left(a+c\right)}+\frac{1}{4\left(b+a\right)\left(b+c\right)}+\frac{1}{4\left(c+a\right)\left(c+b\right)}\)

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

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

Giờ ta cần chứng minh

\(\frac{a+b+c}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{9}{16\left(ab+bc+ca\right)}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

Ta có:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

Vậy ta có ĐPCM