#10 Hard Inequality Problems

By Cauchy, \[\sum\frac{a^2}{1+a}\ge\frac{(a+b+c)^2}{(1+a)+(1+b)+(1+c)}=\frac{9}{6}=\frac32.\] Equality occurs at \(a=b=c=1\).

For \(x\ge0\), the function \(f(x)=1/(x^2+x+1)\) is convex. Jensen gives \[\frac13\sum f(a)\ge f\left(\frac{a+b+c}{3}\right)=f(1)=\frac13.\] Multiplying by \(3\), we get the result.

Let \(q=ab+bc+ca\). Then \(\sum a^2\ge q\), so \((\sum a^2)^2\ge q^2\). Applying \((x+y+z)^2\ge3(xy+yz+zx)\) to \(x=ab\), \(y=bc\), \(z=ca\), gives \(q^2\ge3abc(a+b+c)\).

The inequality is symmetric. By UVW, it is enough to check \(b=c\) and the boundary. For \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\). For \(c=0\), we get \(a^4+b^4-a^3b-ab^3=(a-b)^2(a^2+ab+b^2)\ge0\). Hence the inequality is proved.

By Cauchy, \[\sum\frac{a^2}{b^2+c^2}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)}.\] Since \((a+b+c)^2\ge3(a^2+b^2+c^2)\), the lower bound is \(3/2\).

Let \(q=ab+bc+ca\), \(r=abc\). The left side equals \(9-q\). With \(a+b+c=3\), we have \(q\le3\), so \(9-q\ge6\). Also \(r\le1\), hence \(6r\le6\). Therefore \(9-q\ge6\ge6r\).

We have \((a+b)(b+c)(c+a)=(3-c)(3-a)(3-b)\). The sum of the factors is \(6\). By AM-GM, the product is at most \((6/3)^3=8\). Equality holds at \(a=b=c=1\).

Let \(p=a+b+c\), \(q=3\), \(r=abc\). From \(p^2\ge3q\), \(p\ge3\). From \(q^2\ge3pr\), we get \(9\ge3pr\), so \(r\le3/p\le1\).

Then \(b+c-a=2x\), \(c+a-b=2y\), \(a+b-c=2z\). The sum is \(\frac12\sum\frac{y+z}{x}\). Inside are three pairs of the form \(\frac{y}{x}+\frac{x}{y}\), each at least \(2\), so the sum is at least \(3\).

The right side is \(\sum_{\mathrm{sym}}a^3b\). Thus the statement is Schur degree \(4\), proved via UVW: it is enough to check \(b=c\) and the boundary, where nonnegative squared factors appear.

By Jensen for the concave function \(\sqrt{x}\), \(\frac13\sum\sqrt{a}\le\sqrt{(a+b+c)/3}=1\). Hence \(\sum\sqrt{a}\le3\). The right side of the problem is at least \(3\), since the added sum of squares is nonnegative.

\((1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+abc\). With \(abc=1\), we have \(a+b+c\ge3\) and \(ab+bc+ca\ge3\). Hence the product is at least \(1+3+3+1=8\).

By Cauchy, \[\sum_{\mathrm{cyc}}\frac{a^2}{a^2+ab+b^2}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] It remains to note that \[2(a+b+c)^2-\bigl(2(a^2+b^2+c^2)+ab+bc+ca\bigr)=3(ab+bc+ca)\ge0.\] Hence the denominator is at most \(2(a+b+c)^2\), and the whole fraction is at least \(\frac12\).

By AM-GM, \[2a^4+b^4+c^4\ge4\sqrt[4]{a^8b^4c^4}=4a^2bc.\] Similarly, \[2b^4+c^4+a^4\ge4ab^2c,\qquad 2c^4+a^4+b^4\ge4abc^2.\] Adding the three estimates gives \[4(a^4+b^4+c^4)\ge4abc(a+b+c),\] which proves the claim.

By Schur degree \(3\), \(p^3-4pq+9r\ge0\). For \(p=3\), this is \(27-12q+9r\ge0\). Also \(\sum a^3+6r-q=(27-9q+3r)+6r-q=27-10q+9r\). Since \(q\le3\), \(27-10q+9r\ge27-12q+9r\ge0\). The result follows.

By Cauchy, \(\sum\frac{a^2}{a+b+1}\ge\frac{(a+b+c)^2}{(a+b+1)+(b+c+1)+(c+a+1)}=\frac{9}{9}=1\). Also \(\sum\frac1{a+b+1}\ge\frac{9}{9}=1\). Adding gives the result.

For \(u,v>0\), prove \((v-u)\sqrt{u^2+3v^2}\ge(v-u)(u+v)=v^2-u^2\). If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\), since \(u^2+3v^2-(u+v)^2=2v(v-u)\ge0\). If \(v

From the condition at \(x=0\), we get \(C>0\). The first trinomial is positive for all \(x\), so its discriminant is negative: \[(B-C)^2-4AC<0.\] The discriminant of the second trinomial is \[(B+C)^2-4C(A+B)=B^2-2BC+C^2-4AC=(B-C)^2-4AC<0.\] Its leading coefficient is \(C>0\), hence the second trinomial is also positive for all \(x\).

Since \(n-2

Order the numbers as \(v_1<\cdots